Let $f(t)$ be a real-valued differentiable function on $(-1,1)$ such that $f(0) = 0$ and $ \left| \frac{df}{dt} \right| < 1 $ for $0<t<1$. Then the series $ \sum_{n=0}^{\infty} f(0.5)^n $
Function Analysis Using Mean Value Theorem
We are given a differentiable function $f(t)$ on $(-1,1)$ with $f(0) = 0$ and the condition $ \left| \frac{df}{dt} \right| < 1 $ for $0<t<1$.
By the Mean Value Theorem applied to $f(t)$ on the interval $[0, 0.5]$, there exists a $c \in (0, 0.5)$ such that:
$ f'(c) = \frac{f(0.5) - f(0)}{0.5 - 0} $
Since $f(0) = 0$, the equation simplifies to $ f'(c) = \frac{f(0.5)}{0.5} $. This implies $ f(0.5) = 0.5 \times f'(c) $.
Bound Calculation for $f(0.5)$
The given condition $ \left| \frac{df}{dt} \right| < 1 $ means $ |f'(c)| < 1 $.
Using the derived relationship $ f(0.5) = 0.5 \times f'(c) $, we get:
$ |f(0.5)| = |0.5 \times f'(c)| = 0.5 \times |f'(c)| $.
Because $ |f'(c)| < 1 $, it follows that $ |f(0.5)| < 0.5 \times 1 $, which simplifies to $ |f(0.5)| < 0.5 $.
Series Convergence Test
The series is $ \sum_{n=0}^{\infty} f(0.5)^n $. This is a geometric series with the common ratio $ r = f(0.5) $.
A geometric series $ \sum_{n=0}^{\infty} r^n $ is known to converge absolutely if its common ratio satisfies $ |r| < 1 $.
Convergence Determination
We have established that $ |f(0.5)| < 0.5 $.
Since $ 0.5 < 1 $, the condition $ |f(0.5)| < 1 $ is satisfied.
Consequently, the geometric series $ \sum_{n=0}^{\infty} f(0.5)^n $ converges absolutely.
Important Questions from Mean Value Theorem
A series expansion for the function sin θ is
If f is the derivative of some function on [a, b], then there exists a number c in (a, b) such that Integral of f with respect to x =
Which condition is not required in checking for Taylor's theorem?
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According to the Mean Value Theorem, for a continuous function f(x) in the interval [a, b], there exists a value ξ in this interval such that \(\mathop \smallint \limits_a^b f\left( x \right)dx =\)
