Which condition is not required in checking for Taylor's theorem?
f is necessarily bounded
Understanding Taylor's Theorem and Its Conditions
Taylor's Theorem is a fundamental result in calculus that approximates a function near a point using a polynomial whose coefficients depend on the function's derivatives at that point. It also provides a formula for the remainder term, which quantifies the error in this approximation.
To apply Taylor's Theorem, specifically the version with the Lagrange or Cauchy remainder, certain conditions on the function's continuity and differentiability must be met over a given interval.
Required Conditions for Taylor's Theorem (Lagrange Remainder)
A common statement of Taylor's Theorem with the Lagrange remainder is:
Let $f$ be a real-valued function on $[a, b]$. If
- $f^{(n-1)}$ is continuous on the closed interval $[a, b]$, and
- $f^{(n)}$ exists on the open interval $(a, b)$,
then for each $x \in [a, b]$, there exists a real number $c$ strictly between $a$ and $x$ such that
\begin{equation*} f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots + \frac{f^{(n-1)}(a)}{(n-1)!}(x-a)^{n-1} + \frac{f^{(n)}(c)}{n!}(x-a)^n \end{equation*}
The last term is the remainder term, $R_n(x) = \frac{f^{(n)}(c)}{n!}(x-a)^n$.
Analyzing the Given Options
Let's examine each option in light of the required conditions for Taylor's theorem:
- It's (n -1)th derivative is derivable on (a, b): If the $(n-1)^{\text{th}}$ derivative, $f^{(n-1)}$, is derivable on $(a, b)$, this means its derivative, which is $f^{(n)}$, exists on $(a, b)$. This is explicitly condition 2 above. Thus, this is a required condition (or a direct consequence of one).
- It's (n - 1)th derivative is continuous on [a, b]: This is explicitly condition 1 above. Thus, this is a required condition.
- f is necessarily bounded: If $f^{(n-1)}$ is continuous on $[a, b]$, then $f^{(n-2)}$ is continuous on $[a, b]$, and continuing this process down to $f^{(0)} = f$, we conclude that $f$ is continuous on $[a, b]$. A fundamental theorem in real analysis states that a continuous function on a closed and bounded interval $[a, b]$ is necessarily bounded on that interval. Therefore, the boundedness of $f$ on $[a, b]$ is a *consequence* of the required condition that $f^{(n-1)}$ is continuous on $[a, b]$. It is not an independent condition that must be checked before applying the theorem; it's a property the function will have if the other conditions hold.
- there exists a real number between a and b: The theorem *guarantees* the existence of a real number $c$ strictly between $a$ and $x$ (and since $x$ is in $[a,b]$, $c$ will be between $a$ and $b$ unless $x=a$). This point $c$ is where the $n^{\text{th}}$ derivative is evaluated in the remainder term. The existence of such a point $c$ is part of the theorem's conclusion, not a condition you must verify beforehand to apply the theorem.
Identifying the Condition Not Required
Based on the analysis, options 1 and 2 describe conditions directly related to the differentiability and continuity requirements necessary to apply Taylor's Theorem. Option 4 describes a result guaranteed by the theorem (existence of point c). Option 3 states that $f$ is necessarily bounded. While $f$ *will* be bounded if the continuity conditions on $[a,b]$ are met, its boundedness is a derived property, not an independent requirement to check upfront. Therefore, the boundedness of $f$ is not a condition that is *required to be checked* separately when checking for Taylor's theorem; it follows from the other requirements.
Thus, the condition that is not required in checking for Taylor's theorem is that $f$ is necessarily bounded.
Revision Table: Taylor's Theorem Conditions
| Condition Mentioned | Is it a Required Condition to Check? | Explanation |
|---|---|---|
| (n-1)th derivative is derivable on (a, b) ($f^{(n)}$ exists on (a, b)) | Yes | This is a standard requirement for the $n^{\text{th}}$ derivative in the open interval. |
| (n-1)th derivative is continuous on [a, b] ($f^{(n-1)}$ is continuous on [a, b]) | Yes | This is a standard continuity requirement on the closed interval. |
| f is necessarily bounded | No (it's a consequence) | If $f^{(n-1)}$ is continuous on [a, b], then $f$ is continuous on [a, b], and thus bounded on this closed interval. |
| There exists a real number between a and b (the point 'c') | No (it's a result) | The theorem guarantees the existence of this point in the remainder term, it's not a condition you check before applying the theorem. |
Additional Information: Taylor Series vs. Taylor Polynomial
It's important to distinguish between Taylor Polynomials with a remainder term (which Taylor's Theorem deals with over a finite interval) and infinite Taylor Series. The conditions for convergence of an infinite Taylor series involve checking the behavior of the remainder term as $n \to \infty$. The theorem discussed here provides the basis for the remainder term for a finite Taylor polynomial.
The existence of derivatives up to order $n$ at the point of expansion ($a$) is always necessary for the Taylor polynomial coefficients to be defined ($f^{(k)}(a)$ for $k=0, \dots, n-1$). Taylor's Theorem with remainder provides conditions over an interval $[a,b]$ or $(a,b)$ to characterize the error when approximating $f(x)$ with the Taylor polynomial.
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