Trying to solve $f(x)f(y) = f(x + y)$ Let $f$ be a differentiable function $f: \mathbb{R} \mapsto \mathbb{R}$ such that   $\forall x, y \in \mathbb{R}, \, f(x + y) = f(x)f(y)\tag{1}$  If $y=0$ then $f(x + 0) = f(x)f(0)$ which is just $f(x) = f(x)f(0)$. Then if $f(0) = 0$, we have $f(x) = f(x)f(0) = f(x) \cdot 0 = 0$ for all $x$.   If $f(0) \neq 0$ instead then $f(0) = f(0 + 0) = f(0)f(0) = f(0)^2$ implies that $0 = f(0)^2 - f(0) = f(0)(f(0) - 1)$ which has solutions $f(0) = 0$ or $f(0) = 1$, but since we assume $f(0) \neq 0$, it must be the case that $f(0) = 1$.   Furthermore let's assume the possibility of some real $z$ such that $f(z) = 0$. Then for all $x$ we have $f(x) = f(z + (x - z)) = f(z)f(x-z) = 0 \cdot f(x-z) = 0$. In other words, if the function equals $0$ at one point, it is zero for all of them. The contrapositive is that if the function is nonzero for any value, then it does not equal $0$ anywhere else. Therefore in the case where $f(0) = 1$, we have $f(x) \neq 0$ for all $x \in \mathbb{R}$.  Going forward we assume the case where  $f(0) = 1 \tag{2}$  By induction, $f(x)^n = f(nx)$ for all positive integers $n&gt;0$ as follows.   Base case with $n=1$, we have $f(x)^1 = f(1 \cdot x)$ which is $f(x) = f(x)$ which is trivially true.  Inductive step we assume $f(x)^n = f(nx)$ and we must show that $f(x)^{n+1} = f((n+1)x)$.   $\begin{align} f(x)^n &amp;= f(nx) &amp; \text{By inductive hypothesis}\\ f(x)^n f(x) &amp;= f(nx)f(x) \\ f(x)^{n+1} &amp;= f(nx + x) &amp; \text{By equation (1)}\\ f(x)^{n+1} &amp;= f((n+1)x) \end{align}$  Thus for the positive integers we prove the result:  $\forall n \in \mathbb{Z^{+}}, \, f(x)^n = f(nx)\tag{3}$  Next we see that $f(x) = f(\frac{x}{2} + \frac{x}{2}) = f(\frac{x}{2})^2$ and since any real number squared is non-negative, we combine this with our assumption that $f(x) \neq 0$ for all $x \in \mathbb{R}$ and so it follows that   $\forall x \in \mathbb{R}, \, f(x) &gt; 0 \tag{4}$  so we are free to divide by $f(x)$ values safely without concerning ourselves with division by $0$ issues.  Since $1 = f(0) = f(x + (-x)) = f(x)f(-x)$ it follows that  $\forall x \in \mathbb{R}, \, f(x)^{-1} = f(-x) \tag{5}$  Then for some negative integer $n &lt; 0$ we have $f(x)^n = (f(x)^{-n})^{-1} = f(-nx)^{-1} = f(nx)$ by $(3)$ and $(5)$. Therefore:  $\forall n \in \mathbb{Z^{-}}, \, f(x)^n = f(nx)\tag{6}$  We can also directly show that $f(x)^n = f(nx)$ holds for $n=0$, since $f(x)^0 = 1 = f(0 \cdot x) = f(0)$ which holds by assumption $(2)$. Now we can conclude:  $\forall n \in \mathbb{Z}, \, f(x)^n = f(nx)\tag{7}$  Now if we have $n \neq 0$ we can take the $n$th root of both sides of equation $(7)$ and get $f(x) = f(nx)^{1/n}$ which also implies that $f(x/n) = f(x)^{1/n}$.   Then for integers $m, n$ with $n \neq 0$ we have $f(m \frac{x}{n}) = f(mx)^{1/n}$ using this result, and then applying equation $(3)$ we transform it to $(f(x)^m)^{1/n}$ or $f(x)^{m/n}$ (again for $n \neq 0$).  If we let some rational $r = m/n$ then we see that  $\forall r \in \mathbb{Q}, \forall x \in \mathbb{R}, f(rx) = f(x)^r \tag{8}$  By setting $x=1$ it is revealed that   $\forall r \in \mathbb{Q}, \, f(r) = f(1)^r \tag{9}$  And then there's some theory I don't understand that I can apply here when $f$ is continuous (as assumed up front since we say $f$ is differentiable) that lets me jump to  $\forall x \in \mathbb{R}, \, f(x) = f(1)^x \tag{10}$  Is my proof right so far?
continuous functions on $\mathbb R$ such that $g(x+y)=g(x)g(y)$ Let $g$ be a function on $\mathbb R$ to $\mathbb R$ which is not identically zero and which satisfies the equation $g(x+y)=g(x)g(y)$ for $x$,$y$ in $\mathbb R$.   $g(0)=1$. If $a=g(1)$,then $a&gt;0$ and $g(r)&gt;a^r$ for all $r$ in $\mathbb Q$.   Show that the function is strictly increasing if $g(1)$ is greater than $1$, constant if $g(1)$ is equal to $1$ or strictly decreasing if $g(1)$ is between zero and one, when $g$ is continuous.
Restoring backups failing Ubuntu 16.04 What's the difference between iterating over a file with foreach or while in Perl? If $\forall x \in R, x^2-x \in Z(G)$, than $R$ is commutative Technical term for associating a group with the actions of a few Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$ Prove if $|z| &lt; 1$ and  $ |w| &lt; 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| &lt; 1$Given that $|1-zw^*|^2 - |z-w|^2 = (1-|z|^2)(1-|w|^2)$I think the first part can be proven by saying $|1-zw^*| = 0$ if and only if $zw^*$ = 1.  And given the conditions that cannot be true. However I don't know if this part is right. Fix missing ssl-cert for subdomain www. using redirect or similar VBA Hide Rows that contain N/A How does $cos(x) = \frac{\vec{v} \cdot \vec{w}}{|\vec{v}| \cdot |\vec{w}|}$ make sense? In Multivariable Calculus, the professor said that in order to compute the angle $x$ between two vectors $v$ and $w$, we use the formula: $\cos(x) = \frac{\vec{v} \cdot \vec{w}}{|\vec{v}| \cdot |\vec{w}|}$  He also mentioned "Projection" but I didn't understand much of it. The only thing I got was that we can create a right triangle by drawing a perpendicular line from one vector to the head of the other and compute the angle like that using $\cos$, but I don't understand why we're multiplying vectors, or dividing by their magnitudes…how do those operations make sense?  Also, in the specific example, the result of $\cos(x) = \frac{\vec{v} \cdot \vec{w}}{|\vec{v}| \cdot |\vec{w}|}$ ended up being $1/2 = \pi/2$. Where does $\pi/2$ come from here? I can't understand how we go from $1/2$ to $\pi/2$. Inner Product Spaces I am trying to reconcile the definition of Inner Product Spaces that I encountered in Mathematics with the one I recently came across in Physics.  In particular, if $(,)$ denotes an inner product in the vector space $V$ over $F$:    $(u + v, w) = (u, w) + (v, w) \text{ for all } u, v, w \in V$, $(\alpha v, w) = \alpha(v, w) \text{ for all } v, w \in V$ and $\alpha \in F$, $(v, w) = (w, v)^* \text{ for all } v, w \in V$, (* denotes complex conjugation)   were some of the properties listed in my mathematics course.    In physics, however, the inner product was said to be linear in the second argument and $(v,\sum(\lambda_i w_i)) = \sum(\lambda_i (v,w_i))$ where $v$ and $w_i$ are kets in Hilbert space and $\lambda_i$ are complex numbers.    To me, these properties of an Inner Product are not compatible.  If the first definition of inner product is correct, then I think $(v,\sum(\lambda_i w_i)) = \sum((\lambda_i)^* (v,w_i))$ where $^*$ denotes complex conjugation. I have noticed that while asking questions and scrolling to the bottom of the page, the right sidebar containing recent questions covers the footer partially. Also, a part of it goes beyond the browser's boundary. It may not be a big problem but it still doesn't look good.    This also happens on the  page I’m having trouble finding an official clarification of this, so I’m wondering if anyone has found a ruling:  Solarians gain powers known as stellar revelations. These are either “photon” or “graviton” revelations. It is specifically mentioned they can be used out of combat, though most include improved effects or separate powers that can only be used “while attuned” - a state a Solarian enters during combat. They may be photon or graviton attuned, at the player’s choice, and are able to increase their level of attunement each round until they become fully attuned to one or the other. However, apart from the very powerful “zenith revelations”, nowhere in the rules mentions whether photon revelations may only be used while photon attuned, or graviton revelations only while graviton attuned. The revelations themselves usually say “while attuned or fully attuned” for the more powerful effects, without mentioning the mode, and there’s no general rule in the class feature that answers this question.  Is there an official ruling on whether the “only while attuned” power of a photon revelation can be used when the character is graviton attuned, or vice versa? I suspect the answer is yes but it’s hard to tell based on the core rules. pwd without symlinks Showing that $e^x > 1+x$ Captured variable in a loop in C# I met an interesting issue about C#. I have code like below.  List&lt;Func&lt;int&gt;&gt; actions = new List&lt;Func&lt;int&gt;&gt;();  int variable = 0; while (variable &lt; 5) {     actions.Add(() =&gt; variable * 2);     ++ variable; }  foreach (var act in actions) {     Console.WriteLine(act.Invoke()); }   I expect it to output 0, 2, 4, 6, 8. However, it actually outputs five 10s.  It seems that it is due to all actions referring to one captured variable. As a result, when they get invoked, they all have same output.  Is there a way to work round this limit to have each action instance have its own captured variable? How to assign same Data to multiple object copies I made many copies of an object, then I thought it would've been better if I made them to share the same Data (being an instance of the same original object), I know how to do it one by one, is there a way to do it for all of them at once? I tried Copy Attribute add-on but didn't find what option can do that, or even if this is possible in the first place. Index entries duplicated for captions How to remove duplicate values from a multi-dimensional array in PHP How can I remove duplicate values from a multi-dimensional array in PHP?  Example array:  Array (     [0] =&gt; Array     (         [0] =&gt; abc         [1] =&gt; def     )      [1] =&gt; Array     (         [0] =&gt; ghi         [1] =&gt; jkl     )      [2] =&gt; Array     (         [0] =&gt; mno         [1] =&gt; pql     )      [3] =&gt; Array     (         [0] =&gt; abc         [1] =&gt; def     )      [4] =&gt; Array     (         [0] =&gt; ghi         [1] =&gt; jkl     )      [5] =&gt; Array     (         [0] =&gt; mno         [1] =&gt; pql     )  ) What are FML and BWSS? I see this ad sometimes and I would like to know what the acronyms are. jQuery autocomplete tagging plug-in like StackOverflow's input tags? When is using the past perfect tense not necessary? Is it more natural to speak in simple past or past perfect when explaining past events to a friend? It seems like Americans use more simple past in everyday life than past perfect. I found this example online:     I had almost completed my essay when my computer suddenly crashed and I noticed that I had forgotten to save the document.   Is it wrong to say the following instead?     I was almost done with my essay when my computer suddenly crashed and I noticed that I forgot to save the document.   Maybe it's just my friends, but I rarely hear them speak in past perfect, so I'm really torn about using past perfect when it isn't necessary.