Showing posts with label calculus help. Show all posts
Showing posts with label calculus help. Show all posts

Sunday, May 24, 2009

Question on Limits and Convergence of Number Sequence

In Number theory tutorialoffered by TutorVista will help you to solve problem related to number sequences, calculating limits and continuity of the sequence.

Topic : Limits and convergence of given number sequence.

Decrease in the value of the numbers in the sequence is said to be sequence convergence.

Question : Write the first five terms of the sequence, determine whether the sequence converges, and if so find its limit.

(1 - \frac{1}{2}), (\frac{1}{2} - \frac{1}{3}), (\frac{1}{3} - \frac{1}{4}), (\frac{1}{4} - \frac{1}{5}), . . . . . .



Solution :


General term will be given as ,
(\frac{1}{n}-\frac{1}{n+1})<br />\\ a_ =  \frac{1}{1} - \frac{1}{1+1} = 1 - \frac{1}{2} = \frac{2-1}{2} = \frac{1}{2} = 0.5 \\ a_2 =  \frac{1}{2} - \frac{1}{3} =  \frac{3-2}{2*3} = \frac{1}{6} = 0.167 \\a_3=  \frac{1}{3} - \frac{1}{4} =  \frac{4-3}{3*4} = \frac{1}{12} = 0.083 \\ a_4 =  \frac{1}{4} - \frac{1}{5} =  \frac{5-4}{5*4} = \frac{1}{20} = 0.05 \\ a_5 = \frac{1}{5} - \frac{1}{6} =  \frac{6-5}{6*5} = \frac{1}{30} = 0.03


















So 0.5, 0.167, 0.083, 0.05, 0.03 Thus we can observe that the values goes on decreasing.
Hence the number sequence converges.

Nows lets find it's limit by ratio test

\mathop{\lim}\limits_{n \to \infty}(\frac{a_{n+1}}{a_n}) \\ = \mathop{\lim}\limits_{n \to \infty}(\frac{\frac{1}{n+1} -\frac{1}{n+1+1}}{\frac{1}{n} -\frac{1}{n+1}}) \\ = \mathop{\lim}\limits_{n \to \infty}(\frac{\frac{n+2-(n+1)}{(n+1)(n+2)}}{\frac{n+1-n}{n(n+1)}}) \\ \mathop{\lim}\limits_{n \to \infty} (\frac{n+2-n-1}{(n+1)(n+2)} \div \frac{n+1-n}{n(n+1)}) \\ \mathop{\lim}\limits_{n \to \infty}(\frac{1}{(n+1)(n+2)}X\frac{n(n+1)}{1}) \\ \mathop{\lim}\limits_{n \to \infty}(\frac{n}{n+2}) \\ = \frac{\infty}{\infty+2} \\ = \frac{\infty}{\infty} \\ = 1<br />

















For more help contact math help or calculus help.

Thursday, May 7, 2009

Question on Finding Volume of the Solid, Using Shell Method for setup and Evaluating the integral

Here is a Post on Finding Volume of a Solid by shell Method set up and Evaluating the integral.

Topic : Volume of a Solid

On Integration also one can find out volume of any solid.

Question : Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis x + y2 = 16

Solution :
We know that the volume of the solid formed by rotating the area between the curve of f(y)
and the lines y = a and y = b about the x-axis is given by,




Given curve is x + y2 = 16 Or x = 16 - y2
The graph of the line “x = 16 - y2” between the two axes is denoted by the shaded region in the graph drawn below:

















So, the volume of the shaded region when revolved around x-axis is given by:



















Hence, the volume of the solid generated by revolving the curve “x + y2 = 16” about the x-axis is “128π”.

If you have any queries do write to our calculus help.