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3 years ago

What's the mean, median, mode and midrange

Mathematics

1 answer:

AVprozaik [17]3 years ago

4 0

Mean= average( add all numbers and divide by the amount of numbers)
median=the middle number
mode= the number that occurs the most ( there can be multiple)
midrange= the average of the smallest and largest numbers

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What is 2÷4 I definitely don't know

DochEvi [55]

Answer:

2÷4 is equal to 0.5.

Take the first digit of the dividend from the left. Check if this digit is greater than or equal to the divisor.
Then divide it by the divisor and write the answer on top as the quotient.
Subtract the result from the digit and write the difference below.

Please Mark Brainliest If This Helped!

7 0

1 year ago

Find two linearly independent solutions to the equation y"-2xy'+2y=0 in the form of a power series.

ioda

We want a solution in the form

$y=\displaystyle\sum_{n\ge0}a_nx^n$

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting $y$ and its derivatives into the ODE,

$y''-2xy'+2y=0$

gives

$\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n-2\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}+2\sum_{n\ge0}a_nx^n=0$

Shift the index on the second sum to have it start at $n=1$ :

$\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}=\sum_{n\ge1}na_nx^n$

and take the first term out of the other two sums. Then we can consolidate the sums into one that starts at $n=1$ :

$\displaystyle(2a_2+2a_0)+\sum_{n\ge1}\bigg[(n+2)(n+1)a_{n+2}+(2-2n)a_n\bigg]x^n=0$

and so the coefficients in the series solution are given by the recurrence,

$\begin{cases}a_0=y(0)\\a_1=y'(0)\\(n+2)(n+1)a_{n+2}=2(n-1)a_n&\text{for }n\ge0\end{cases}$

or more simply, for $n\ge2$ ,

$a_n=\dfrac{2(n-3)}{n(n-1)}a_{n-2}$

Note the dependency between every other coefficient. Consider the two cases,

If $n=2k$ , where $k\ge0$ is an integer, then

$k=0\implies n=0\implies a_0=a_0$

$k=1\implies n=2\implies a_2=-a_0=2^1\dfrac{(-1)}{2!}a_0$

$k=2\implies n=4\implies a_4=\dfrac{2\cdot1}{4\cdot3}a_2=2^2\dfrac{1\cdot(-1)}{4!}a_0$

$k=3\implies n=6\implies a_6=\dfrac{2\cdot3}{6\cdot5}a_4=2^3\dfrac{3\cdot1\cdot(-1)}{6!}a_0$

$k=4\implies n=8\implies a_8=\dfrac{2\cdot5}{8\cdot7}a_6=2^4\dfrac{5\cdot3\cdot1\cdot(-1)}{8!}a_0$

and so on, with the general pattern

$a_{2k}=\dfrac{2^ka_0}{(2k)!}\displaystyle\prod_{i=1}^k(2i-3)$

If $n=2k+1$ , then

$k=0\implies n=1\implies a_1=a_1$

$k=1\implies n=3\implies a_3=\dfrac{2\cdot0}{3\cdot2}a_1=0$

and we would see that $a_{2k+1}=0$ for all $k\ge1$ .

So we have

$y(x)=\displaystyle\sum_{k\ge0}\bigg[a_{2k}x^{2k}+a_{2k+1}x^{2k+1}\bigg]$

so that one solution is

$\boxed{y_1(x)=\displaystyle a_0\sum_{k\ge0}\frac{2^k\prod\limits_{i=1}^k(2i-3)}{(2k)!}x^{2k}}$

and the other is

$\boxed{y_2(x)=a_1x}$

I've attached a plot of the exact and series solutions below with $a_0=y(0)=1$ , $a_1=y'(0)=1$ , and $0\le k\le5$ to demonstrate that the series solution converges to the exact one.