Recently moved from Canon to Sony a7iii. My wife has pointed out that a lot of the skin tones look much more red than previously.  Any suggestions on avoiding this?  I've seen there is a slider to adjust white-balance but would prefer not having to manually guess/set this for every photo.  Does the Sony have skin specific settings or picture styles that might help?
I have close to two hundred images where the subject looks like he just walked out of a sauna. It was 36 degrees with humidex, but he was not that red.   For comparison, attaching another picture, same camera, same lens, auto white balance, next day.       So, two questions, 1) how did that happen?, and 2) how do I fix it? I can bring down the red, but then I have to manually bring back the lip color in 200 pictures.   And next time when I deal with an older guy at 36 degrees C I will be shooting indoors.   *Edit: The camera is Olympus EPM 2. I am now guessing that he may have been THAT red, but my eyes tricked me. The last picture is from the beginning of the shoot before he got very hot and had two glasses of wine AND the background worked better for his skin and clothes. I do not have colour management setup, but I will get on it.
Will database replication increase the size of the replicated database? First thank you for your time. We have a SQL Server database that was assigned to be replicated. the original size of the database was 5 GB for the datafile, not the log file. Now after the replication the data file is 4 times bigger around 19 GB. Any pointers to see what could happened will be appreciated. I know sql as developer and some little management but nothing about replication. The performance of the database has been damaged very bad so I am trying to see what need to be done to improve it. I do apologize for my English as well. In a family with two children, what are the chances, if one of the children is a girl, that both children are girls? Replace HTML page with contents retrieved via AJAX I have an HTML page with a typical structure:  &lt;html&gt;   &lt;head&gt;    &lt;script src="..." &gt;&lt;/script&gt;    &lt;style&gt;...&lt;/style&gt;   &lt;/head&gt;   &lt;body&gt;    content   &lt;/body&gt;   &lt;script&gt;     var success_callback = function(data) {       // REPLACE PAGE CONTENT &amp; STRUCTURE WITH "data"     }     ajax(url, params, success_callback);   &lt;/script&gt; &lt;/html&gt;   Do you think it is possible ?  I've already tried to give the html tag an id and doing $(id).replace(data); with no success.  Don't ask me why, but that is what I need (I'm working with a special "mashup builder" site... it is a long story).  EDIT : I forgot to say that scripts in the received content have to be executed, even external scripts included using &lt;script src="..."&gt;. Proving $\sqrt2$ is irrational I used the method of contradiction by assuming that $\sqrt 2$ is a rational number. Then, by the definition of rational number, there exist two integers $p$ and $q$ whose ratio equals $\sqrt 2$. Thus, $$\frac pq = \sqrt2\tag{x}$$  Squaring both sides, $$p^2/q^2 = 2\tag{a}$$  or $$p^2 = 2q^2\tag{b}$$ This means that $p^2$ is an even number, implying that $p$ is even.  Now, any even integer can be written as $2^kf$ where $f$ is any odd integer and $k$ is some positive integer (the minimum value of $k$ and $f$ is $1$ since even numbers start from $2 = 2^1\cdot1$). For odd numbers, $k=0$.  For example, $8 = 2^3 \cdot 1$, $4= 2^2 \cdot 1$, $18= 2^1 \cdot 9$, $24 = 2^3 \cdot 3$, $-12= 2^2 \cdot(-3)$, etc.  Now, from $(b)$, $q^2$ can be even or odd.  Case 1: $q^2$ is even (thus meaning $q$ is even). Then $p= 2^{k_1} \cdot f_1$ (say) and $q = 2^{k_2} * f_2$ (say). Note here in this case, both conditions $k_1=k_2$ and $f_1=f_2$ can't hold simultaneously since that would mean $p/q =1$ and here $p/q =\sqrt2$ (from $(x)$)). Let's consider the condition when $k_1=k_2=k$ but $f_1\ne f_2$. Then $$\frac{p^2}{q^2}=\frac{(2^kf_1)^2}{(2^kf_2)^2} = \frac{f_1^2}{f_2^2} =2$$ (from $(a)$), i.e.: odd/odd ($f_1$ and $f_2$ are odd) can never equal $2$.  Now lets consider $f_1=f_2=f$ but $k_1\ne k_2$, thus $(2^{k_1}f)^2/(2^{k_2}f)^2= 2^{2(k_1-k_2)}= 2$ meaning $k_1 - k_2 = 0.5$ but $k_1,k_2$ are integers, so their difference can't be $0.5$. (Also here, $k_1-k_2$ must be greater or equal to $1$ since $2^{k_1-k_2}=\sqrt 2$, $k_1-k_2$ can't be negative since $\sqrt 2&gt;1$, but $k_1-k_2\ge1$ can't satisfy $(a)$ since the minimum value of $p/q$ in this case will be $2$ which is surely greater than $\sqrt2$.)  Now lets consider $f_1\ne f_2$ and $k_1\ne k_2$, thus $\frac{(2^{k_1} f_1)^2}{(2^{k_2} f_2)^2}= 2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal $2$ since $\frac{f_1^2}{f_2^2}$ is either odd or "odd/odd" ie: it doesn't contain 2 as a factor and $2^{2(k_1-k_2)}$ is a power of $2$. So in $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$, there's no chance of cancellation of the odd factor $\frac{f_1^2}{f_2^2}$.  Case 2: $q^2$ is odd (thus meaning $q$ is odd). $q=f'$ (say, here $k=0$ for $q$ but not for $p$), thus from $(a)$ $2^{2k}\frac{f^2}{f'^2} = 2$ but this isn't possible since a power of $2$ multiplied with odd factor can't equal $2$.  Thus, both case 1 and 2 suggest that for any possible combination of $k_1,k_2,f_1$ and $f_2$, $p/q\ne\sqrt2$, i.e.: for no value of integers $p$ and $q$, $p/q = \sqrt2$. Thus, this contradicts our assumption that $\sqrt2$ is rational. Therefore it must be irrational.  PLS NOTE that i would like to clarify that $f_1^2$/$f_2^2$ is either an odd integer or a fraction of odd/odd or 1/odd form, thus not containing  2 as a factor,for any $2^n$, if it has to be reduced to 2 , must be multiplied with $1/ 2^(n-1)$ but $f_1^2$/$f_2^2$ can't be of $1/2^[n-1]$ form since for odd nos, in $2^k$*f notation, k is 0 , so 2 vanishes as $2^k$ becomes 1 in this case thus f1/f2 can't be represented as  1/ $2^(n-1)$ in which n has to be existent.  therefore $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal 2 as to further explain the 3rd condition of case 1  Difficulty: is my approach correct? This proof which I thought is different from proofs found on the internet or books since I have used $p$ and $q$ as any integers, which may have a common factor. So I am not sure if I am on right track. Will someone please check this out and make kind comments? Proof of the inequality $(x+y)^n\leq 2^{n-1}(x^n+y^n)$ Where can I find every gnome? 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