Why In the Taylor's remainder theorem they used "$\equiv$" sign?

Asked Dec 24 '20 at 19:19

Active Dec 24 '20 at 19:30

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Assume that $f(x)$ has $n+1$ continuous derivatives on an interval $\alpha\le x \le\beta$, and let the point $a$ belong to the interval. For nth degree Taylor polynomial $P_n(x)$ let $R_n(x)\equiv f(x)-P_n(x)$ denote the remainder in approximating $f(x)$ by $P_n(x)$. Then

$$R_n(x)=\frac{(x-a)^{n+1}}{(n+1)!}f^{(n+1)}(c_x), \quad \alpha\le x\le \beta$$with $c_x$ an unknown point between $a$ and $x$.

I don't know why they wrote $R_n(x)\color{red}{\equiv}f(x)-P_n(x)$.

edited Dec 24 '20 at 19:30

asked Dec 24 '20 at 19:19

Etemon

6,437

6

Sometimes that symbol is used to mean "defined to be equal to." – B. Goddard Dec 24 '20 at 19:21
What do you mean by $\equiv$. – hamam_Abdallah Dec 24 '20 at 19:21
I asked a similar question of sorts a while back here. But basically it can often be used to denote definitions or identities; you can think of it as an equal sign underlined for emphasis (intuitively, of course, not literally). – PrincessEev Dec 24 '20 at 19:23
@B.Goddard but here isn't the right side and left side equal ? why we defined to be equal? – Etemon Dec 24 '20 at 19:23
@EeveeTrainer I read about their differences before but here I don't understand whether it means it is an identity or something else. – Etemon Dec 24 '20 at 19:25
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In this case, it means we're defining the object "$R_n(x)$" to be equal to $f(x) - P_n(x)$, i.e. the error in approximating $f(x)$ by $P_n(x)$. – PrincessEev Dec 24 '20 at 19:26
Oh I got it thank you. – Etemon Dec 24 '20 at 19:27
@hamam_Abdallah it is identity sign. – Etemon Dec 24 '20 at 19:29

Why In the Taylor's remainder theorem they used "$\equiv$" sign?

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