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I had the equation:

$$\frac{1}{(1+x)(1-x)^2}=\frac{A}{(1-x)^2}+\frac{B}{1-x}+\frac{C}{1+x}$$

$$A=\left.\frac{1}{1+x}\right|_{x=1} = \frac{1}{2} \tag2$$

What does $(2)$ mean? Particularly, the notation $|_{x=1}$

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4 Answers 4

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Short Version

The notation $$ \left. \frac{1}{1+x} \right|_{x=1} $$ means (essentially) "evaluate the expression $1/(1+x)$ when $x = 1$". This is equivalent to restricting the implicitly defined function $x \mapsto 1/(1+x)$ to the domain $\{ 1 \}$, and then determining the the image of that function.

In More Detail

This can be seen as a special case of the restriction of a function to a smaller domain. In general, if $f : X \to Y$ and $A \subseteq X$, then we may define a new function, denoted by either $f|_A$ or $f|A$, via the formula $$ f|_A : A \to Y : x \mapsto f(x). $$ This new function is called the restriction of $f$ to $A$.

This concept often comes up first in elementary classes when one wants to find inverses. For example, the function $x \mapsto x^2$ is not invertible over the entire real line, but if you restrict it to the interval $(-\infty, 0]$, it becomes invertible. We get a new function $$ f|_{(-\infty,0]} : (-\infty,0] \to \mathbb{R}$$ which is defined by $$ f|_{(-\infty, 0]}(x) = x^2 $$ whenever $x \le 0$, and is undefined otherwise. Note that $$ (f|_{(-\infty,0]})^{-1}(y) = -\sqrt{y} $$ for any nonnegative real number.

In the current context, the notation $$ \left. \frac{1}{1+x} \right|_{x=1} $$ can be seen as a special case of the same idea: the formula $1/(1+x)$ defines a function on $\mathbb{R}\setminus\{-1\}$, but we are only interested in the value of this function at $x=1$. We could define $$ f(x) = \frac{1}{1+x}, $$ and then evaluate $f(1)$. Evaluating $f(1)$ is equivalent to restricting $f$ to the domain consisting only of that single point and determining the image of that function, so we only have to consider $$ f|_{\{1\}} = f|_{\{x : x=1\}}. $$ Because all of those curly braces are kind of a pain, we can simplify the notation by writing simply $f|_{x=1}$. Now, $f|_{x=1}$ is a very simple function, which takes only one value, so we might as well just assume that $f|_{x=1}$ denotes this value which, in this case, is $\frac{1}{2}$.

Or, instead of going through all of that, we just write $$f(\color{blue}{x})\big|_{x=1} = \left. \frac{1}{1+\color{blue}{x}} \right|_{x=1} = \frac{1}{1+\color{blue}{1}} = \frac{1}{2}. $$

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$f(x)|_{x=c}$ denotes the function $f(x)$ evaluated at $x=c$, so in particular it is equal to $f(c)$ if $f$ is defined at $x=c$.

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By definition, $f(x)_{|x=1} = f(1)$.

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It means you substitute $x=1$ into the expression. Thus $$A=\left.\frac{1}{1+x}\right|_{x=1} =\frac{1}{1+1} = \frac{1}{2}. $$

This notation can also be used as follows:

$$\int_{0}^{1}\frac{1}{x+1}dx=\ln(x+1)\bigg|_{x=0}^{x=1}=\ln(1+1)-\ln(0+1)=\ln(2).$$

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  • $\begingroup$ Now, I suddenly understand your notations by their context. They are new to me. Thank you for your example. $\endgroup$ Jan 25, 2021 at 9:55
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    $\begingroup$ You're very welcome! :-) $\endgroup$
    – Alessio K
    Jan 25, 2021 at 11:48

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