# $|_{x=1}$ notation?

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

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

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

### 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 $$f(\color{blue}{x})\big|_{x=1} = \left. \frac{1}{1+\color{blue}{x}} \right|_{x=1} = \frac{1}{1+\color{blue}{1}} = \frac{1}{2}.$$

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

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

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:

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

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