please help me with this. It's urgent and I would appreciate any help. Thanks
Hello, simfonija!
You need help with the entire exam?
Should we be helping you?
Having said that, here's #1 . . .
1) Find the interval of convergence of the power series: .$\displaystyle \sum^{\infty}_{n=0}\frac{(1 - \frac{x}{2})^n}{\sqrt{n+1}} $
Ratio Test
$\displaystyle r \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(1 - \frac{x}{2})^{n+1}}{\sqrt{n+2}} \cdot \frac{\sqrt{n+1}}{(1 - \frac{x}{2})^n} \;=\;\sqrt{\frac{n+1}{n+2}}\cdot\left(1 - \frac{x}{2}\right)$
$\displaystyle R\;=\;\lim_{n\to\infty}|r|\;=\;\lim_{n\to\infty}\l eft|\sqrt{\frac{n+1}{n+2}}\cdot\left(1 - \frac{x}{2}\right)\right| \;= \;\left|1 - \frac{x}{2}\right|$ **
For convergence, $\displaystyle R < 1$
So we have: . . . . $\displaystyle \left|1 - \frac{x}{2}\right| \; < \; 1$
Then: . . . . .$\displaystyle -1 \; < \; 1 - \frac{x}{2} \; < \; 1$
Subtract 1: . .$\displaystyle -2 \; < \; -\frac{x}{2} \; < \; 0$
Multiply by -2: . .$\displaystyle 4 \; > \; x \; > \; 0$
Therefore: . . . .$\displaystyle \boxed{0 \: < \: x \: < \: 4}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
We have: .$\displaystyle \lim_{n\to\infty}\sqrt{\frac{n+1}{n+2}}$
Divide top and bottom by $\displaystyle n:\;\;\sqrt{\frac{\frac{n}{n} + \frac{1}{n}}{\frac{n}{n} + \frac{2}{n}}} \;=\;\sqrt{\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}}$
Then: .$\displaystyle \lim_{n\to\infty} \sqrt{\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}} \;= \;\sqrt{\frac{1 + 0}{1 + 0}} \;=\;\sqrt{1} \;= \;1$
Ok, are you playing a teacher and I am bad student??? I was away for more than month (serious family reason) and hoping to manage to pass exam somehow. This is assignment which gives me 3% of total mark. On Monday I ll get copy of the notes. Just help if you can (no provocation please), if can't I also understand... Thanks
I have something to add to Soroban's post.
He did not find the entire interval of convernge.
The endpoints are,
$\displaystyle x=0,x=4$
The two respective series we obtain are,
$\displaystyle \sum_{n=0}^{\infty} \frac{1}{\sqrt{n+1}}$
This can the zeta function $\displaystyle \zeta(1/2)$, alternatively you can use the p-series test for exponent 1/2.
And also, the function,
$\displaystyle f(x)=\frac{1}{\sqrt(x+1)}$ is a positive continous decreasing function and the improper integral:
$\displaystyle \int_0^{\infty} \frac{dx}{\sqrt{x+1}}$ diverges. Thus, $\displaystyle x\not =0$
Checking the point at $\displaystyle x=4$ we have,
$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{n+1}}$
We have that $\displaystyle \lim_{n\to \infty}a_k=0$
The series is strictly decreasing and alternating. Thus, by Leibniz's test it is convergent.
$\displaystyle 0<x\leq 4$
Hi
Task 2 I have no idea how to solve. Task number 3 I tried and I did some calculation but 3 (a) I don’t know why 6^1/2=2(1+1/2)^1/2 is better than the other. For 4 I also don’t have example of such diff equations. I am not so bad in maths but as I said this assignment is for Monday and on Monday I will copy the notes from the colleague.
#4)
Solve, for $\displaystyle x>0$.
$\displaystyle x^2y''-xy'+y=0$.
Since, this is equivalent to,
$\displaystyle y''-\frac{1}{x}y'+\frac{1}{x^2}y=0$ where $\displaystyle p(x),q(x)$ are continous on some open interval there exists a basis of dimension two. Hence two linearly independant solutions.
We are given that $\displaystyle y_1=x$ is a solution.
To find, the second linearly independent solution we use reduction of order (which leads to):
$\displaystyle y_2=y_1\int \frac{W(y_1,y_2)}{y_1^2} dx$
The Wronskian (we can remove the constant multiplier) by Abel's theorem is,
$\displaystyle \mbox{exp} \left( -\int p(x) dx \right)=e^{\ln |x|}=x$
Thus,
$\displaystyle y_2=x\int \frac{x}{x^2}dx=x\ln x$
We see that this is a solution to this differencial equation which is linearly independent to $\displaystyle y_1=x$
Thus, the set of all solutions can be expressed as (general solution),
$\displaystyle C_1 x+C_2 x\ln x$
Congratulation we solved the homogenous equation.
----
Now the non-homogenous equation.
For the purposes of the solution we need to find the Wronskian between these two solutions which is,
$\displaystyle W(y_1,y_2)=\left| \begin{array}{cc} x&x\ln x\\1&1+\ln x \end{array} \right|=x$
Now we use Variation of Parameters.
The formula is,
$\displaystyle -y_1\int \frac{y_2 g(x)}{W(y_1,y_2)}dx+y_2\int \frac{y_1 g(x)}{W(y_1,y_2)}dx$
Thus,
$\displaystyle -x\int \frac{x\ln x\cdot x^{5/2}}{x} dx+x\ln x\int \frac{x\cdot x^{5/2}}{x} dx$
Thus,
$\displaystyle -x\int x^{5/2}\ln x dx+x\ln x\int x^{5/2} dx$
Integrate use by parts on first integral,
$\displaystyle -x\left( \frac{2}{7}\ln x x^{7/2} - \frac{4}{49} x^{7/2} \right) +x\ln x \left( \frac{2}{7}x^{7/2} \right)$
Thus,
$\displaystyle \frac{4}{49}x^{9/2}-\frac{2}{7}x^{9/2}\ln x+\frac{2}{7}x^{9/2}\ln x$
Thus,
$\displaystyle \frac{4}{49}x^{9/2}$
Thus, all solutions can be expressed as,
$\displaystyle C_1 x+C_2 x\ln x+\frac{4}{49}x^{9/2}$