Kit Library / Mathematics / Numerical Analysis
⚡ Topic Learning KitComputational Methods
Shared by a Veda learner · Generated with Veda AI
What you'll study, topic by topic
Computational Methods: Numerical Techniques and Interpolation
~8 min · full explanation, examples & memory tricks in the app
Try the smart MCQs from this kit
25 questions laddered from warm-up to topper-level, each with an explanation. A taste:
Which of the following is the correct statement of Rolle's theorem for a function $f(x)$ on $[a,b]$?
BeginnerShow answer & explanation
$f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$ implies there exists $c \in (a,b)$ such that $f'(c)=0$.
Rolle's theorem states that if a function is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists at least one point $c$ in $(a,b)$ where the derivative is zero.
For the function $f(x) = \frac{\sin x}{e^x}$ on the interval $(0, \pi)$, what does Rolle's theorem guarantee?
IntermediateShow answer & explanation
There exists $c \in (0, \pi)$ such that $f'(c) = 0$.
Since $f(0) = \frac{\sin 0}{e^0} = 0$ and $f(\pi) = \frac{\sin \pi}{e^{\pi}} = 0$, and $f$ is continuous and differentiable on the interval, Rolle's theorem guarantees a point $c$ where $f'(c)=0$.
What is the value of $\Delta^3 (1-x)(1-2x)(1-3x)$ when the interval of differencing is unity?
IntermediateShow answer & explanation
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The third forward difference of a cubic polynomial is constant and equals $6 \times$ leading coefficient $\times h^3$. Here leading coefficient is $-6$, so $\Delta^3 = -36$.
Using the Newton-Raphson method, what is the approximate value of $\sqrt[3]{24}$ correct to three decimal places?
IntermediateShow answer & explanation
$2.884$
Solving $x^3 - 24 = 0$ by Newton-Raphson gives $x \approx 2.884$ after a few iterations.
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