I was doing some exam practice when I almost surely went the wrong way in my method and ended up with the following equation: $$f'(x) = f(x) - f(x-1)$$ Although not what I was intended to find, I'm still curious to know if there are methods to solve this. We can spot the obvious solution $f(x) = Ax+B$ , but I could not find a way to prove this was the only solution, or find any other solutions. Any insight would be helpful. For context I am a first-year undergrad.
The question is:     When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?   The most obvious solution is a linear function of the form $f(x)=ax+b$. Is this the only solution?  Edit  I should add that $f:\mathbb R\to\mathbb R$ to the question.
Is there a power saving application similar to Jupiter? I have already seen a proof of this statement here on stackexchange but I would like someone to verify whether "my proof" is correct and also if you could point out issues with the writing style. thanks in advance!  Statement. $\mathbb{Z}_{p}$ is an integral domain $\Leftrightarrow$ $p$ is prime.  Proof.$(\Rightarrow)$ Suppose $\mathbb{Z}_{p}$ is an integral domain. Assume (by way of contradiction) that $n$ is not prime. Then $n=ab$ for some $a,b\in \mathbb{Z}$ where $1&lt;a$ and $b&lt;n$, meaning $ab\equiv 0\pmod n$. This is a contradiction since $\mathbb{Z}_{p}$ is assumed to be an integral domain and therefore does not contain any zero divisors. Therefore $p$ must be prime.  $(\Leftarrow)$ Assume $p$ is prime. Then $a$ does not divide $n$ for any $1&lt;a&lt;n$ where $a\in \mathbb{Z}_{p}$. Thus $\mathbb{Z}_{p}$ does not contain any zero divisors and is therefore an integral domain. My colleague and I were discussing about certain types of customers in the IT industry. You have to work extremely hard to convince them so that they think about opening their wallets .  However, they should not be mistaken to be stingy (carries a negative connotation). Cost-conscious is a closely related term but is a bit wordy. Another word that I considered is shrewd but I don't know if it fits my context well.  Usage Sentence:     Nowadays, IT customers are becoming increasingly ________. They demand more for less and probably think a zillion times before reaching for their pockets. If they are not convinced 100%, they simply won't open their wallets.    Is there is a noun or an adjective to describe such people? I am also open to using idioms, if there is one that conveys the intentions well.  N.B:  I am looking for words with neutral to positive connotations. Avoiding Windows 10 Creator, but not other updates How to resolve conflicting distributions in apt-get When doing an apt-get update/upgrade I see that warning  W: Conflicting distribution: https://downloads.plex.tv/repo/deb public InRelease (expected public but got )   I was upgrading to Ubuntu Zesty 17.04 before and do not know how to fix the above. Already have seen ? but there are no suggested fixes for that.  Maybe one of you can give some hints? Thanks! When is "to" a preposition and when the infinitive marker? From my (honestly limited) knowledge of what the Heat Death of the Universe is, I understand that HDotU is the point when energy levels across the entire Universe reach equillibrium, thus energy doesn't flow anymore absolutely anywhere, and thus nothing will ever happen anymore.  If that's the case, would time still be? Even when there's not a single event happening inside it and nobody's there to observe the ticking, does time (or space, come to think of it) exist? Or would HDotU cause something that ends the entire spacetime? What is difference between int.class and Integer.TYPE in java? I would like to think that some of the software I'm writing today will be used in 30 years. But I am also aware that a lot of it is based upon the UNIX tradition of exposing time as the number of seconds since 1970.  #include &lt;stdio.h&gt; #include &lt;time.h&gt; #include &lt;limits.h&gt;  void print(time_t rt) {     struct tm * t = gmtime(&amp;rt);     puts(asctime(t)); }  int main() {     print(0);     print(time(0));     print(LONG_MAX);     print(LONG_MAX+1); }   Execution results in:   Thu Jan  1 00:00:00 1970 Sat Aug 30 18:37:08 2008 Tue Jan 19 03:14:07 2038 Fri Dec 13 20:45:52 1901           The functions ctime(), gmtime(), and localtime() all take as an argument a time value representing the time in seconds since the Epoch (00:00:00 UTC, January 1, 1970; see time(3) ).      I wonder if there is anything proactive to do in this area as a programmer, or are we to trust that all software systems (aka Operating Systems) will some how be magically upgraded in the future?  Update It would seem that indeed 64-bit systems are safe from this:  import java.util.*;  class TimeTest {     public static void main(String[] args) {         print(0);         print(System.currentTimeMillis());         print(Long.MAX_VALUE);         print(Long.MAX_VALUE + 1);     }      static void print(long l) {         System.out.println(new Date(l));     } }    Wed Dec 31 16:00:00 PST 1969 Sat Aug 30 12:02:40 PDT 2008 Sat Aug 16 23:12:55 PST 292278994 Sun Dec 02 08:47:04 PST 292269055   But what about the year 292278994? Prove existence of a real root. Problem: If   $$C_0+\frac{C_1}{2}+\cdots + \frac{C_{n-1}}{n}+\frac{C_n}{n+1} =0,$$  where $C_0,...,C_n$ are real constants, prove that the equation  $$C_0+C_1x+\cdots +C_{n-1}x^{n-1}+C_nx^n=0$$  has at least one real root between $0$ and $1$.  Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 4. Fastest way to try all passwords Suppose you have a computer with a password of length $k$ in an alphabet of $n$ letters.  You can write an arbitrarly long word and the computer will try all the subwords of $k$ consecutive letters. What is the smallest word that contains all combinations of $k$ letters as subword? (i.e. the fastest way to hack the computer :) )  The smallest word that contains $n^k$ subwords of size $k$ has length $k-1+n^k$ and based on some easy cases, we would like to prove that it is in fact possible to find a word of such length that contains all possible passwords. The problem can be translated into a problem in graph theory, by taking as vertices all words of length $k$.  We tried $k=2$, where you can prove the conjecture by induction. For $n=2$ and small $k$ it also works. Let $a &lt; b$ and $f:[a, b] \rightarrow \mathbb R$ be a Riemann integral function. Suppose that $f(x) \geq 0$ for all $x \in [a, b]$. I want to show whether $\int_a^bf(x) = 0$ implies $f(x) = 0$ or not, but I'm not sure where to start. I don't know much about Riemann integral functions in general. I only know that $$\underline \int_a^b f= \overline \int_a^b f.$$ Also, I know the fact that if $f$ is continuous, then $f(x) = 0$ for all $x\in [a, b]$. Can anyone provide any hints for me, because I'm really confused about this whole thing. I tried to use the absolute convergence test. $$\int_1^2\left|\frac{\sin{x}}{\log{x}}\right| \, dx&lt;\int_1^2\left|\frac{1}{\log{x}}\right| \, dx$$ but I'm not sure whether $\frac{1}{\log{x}}$ converges or not.  Thanks in advance! var functionName = function() {} vs function functionName() {} There should be infinitely many primes of the form $5+6n$. How do you prove it? The same should be true for $7+6n$. How do I delete an entity from the Minecraft world folder? In my online brokerage account, I want to buy a particular stock and I see the following:  Bid:  13.20 x200           Ask:  13.27 x1,000   the current stock price is 13.22.   Can someone explain what the bid and ask prices mean relative to the current price?  If I buy 1000 shares, why would I pay more? Programmatically delete one entry from one field on one entity What's the proper way to programmatically delete one specific field entry for a multi-entry field from one specific entity? (that's specific fields and specific entities, not field types or instances and entity types)  Here's some possibilities I've looked into which don't work:   Loading the entity with entity_load(), setting the field entry to =null or =array(), then saving with field_attach_update(). It's the closest I've found but it leaves a zombie null field entry in the database which shows up as a blank row on forms until the form is saved and hook_field_is_empty() can kick in (hook_field_is_empty() is configured such that =null or =array() will flag as empty). Loading the entity with entity_load(), unsetting the field entry by key, then saving with field_attach_update(). This seems to do nothing - it seems that Drupal interprets the entry not being there as a sign to not modify it. (strangely I also sometimes get maximum query time expiration timeouts when trying this approach) field_attach_delete() - this is much too blunt: it kills all fields for the entity field_purge_data() - better, but still too blunt: it kills all entries of the field, not specific entries   So to clarify, I've got code that finds (and loads) an entity which has a field that has multiple entries, and finds a specific entry on that field that needs to be removed. I want to remove that entry completely, without touching any other entries or any other fields on the entity. How do I evolve Kadabra to its final form without trading it? INI file parsing in PowerShell