Question

hi,im ronsthel gordula,i'm bs.math student...about the trapezoidal rule and its error.

trpezoidal rule is a technique of approximating definite integrals,with the use of trapezoids and area.

to find the truncation
error...how it is.?

Answer 1

\[\Large \epsilon_{\large{T}}=-\frac{1}{12}(b-a)f''(\eta)(\Delta x)^2,\qquad a <\eta<b\]

Answer 2

can you explain and give me one example for me...?
because all know about the trapezoidal rule is get the value of trapezoid..

Answer 3

example (from the book)
\[\int_0^3 \frac{dx}{16+x^2}\]

Answer 4

\[\Large{f(x)=(16+x^2)^{-1}\\f'(x)=-2x(16+x^2)^{-2}\\f''(x)=(6x-32)(16+x^2)^{-3}}\] we'll find a suitable \(0<\eta<3\) by using an extrema of \(f''(x)\)

Answer 5

\[\Large f'''(x)=24x(16-x^2)(16+x^2)^{-4} > 0 \quad \forall \; x \in [0,3]\]
therefore \[\Large f''(0)=-\frac{1}{128},\qquad f''(3)=\frac{22}{15625}\] are the extrema of \(f''(x)\)

Answer 6

OH I am so not worthy O_O
dat advanced formula *falls down*

Answer 7

\[\int\limits_{0}^{3}(dx/16+x^2)\] if n=4\[\frac{ 3-0 }{ 4}=\frac{ 3 }{ 4 }\]\[Deltax=\frac{ 3 }{ 4}\]\[_{Xo}=0_{}\]\[_{X1}=\frac{ 3 }{ 4 }\]\[_{X2}=\frac{ 3 }{ 2 }\]\[_{X3}=\frac{ 9 }{ 4}\]\[_{X4}=3\] then i substitute the value of \[_{_{X0-X4}}\] to the formula..right?

Answer 8

\[\large -\frac{1}{12}f''(\eta=0)(\Delta x)^2 <\epsilon_{\small {T}}<-\frac{1}{12}f''(\eta=3)(\Delta x)^2 \]
the error is dependent of course, on the number of partitions \(n\) of the interval.

Answer 9

oh, by the way, the error \({\Huge\epsilon}\,_T\) is from numerical analysis.

Answer 10

the formula you are referring to is the trapezoidal rule \[\int_{x_0}^{x_n}f(x)\mathrm dx=\frac{x_n-x_0}{2n}\left[f(x_0)+2f(x_1)+\cdots+2f(x_{n-1})+f(x_n)\right]\]

Answer 11

thank you for helping me until next time. :)

Answer 12

if T is the approximation by tha trapezoidal rule, the error is given by \[{\Huge \epsilon}_T=\int_a^bf(x) dx -T\]
the error can be measured if \(f(x)\) is an integrable function, otherwise, the error can only be approximated by numerical analysis.