What is the general solution for the power series in Calculus 2?

[joku1234]

please help!

 

[chisigma]

please help!

... wellcome on MHB joku1234!...

The regulations state that you've shown some effort to resolve questions ... to get the result you can refer to two standard examples: the geometric and exponential series ...

Kind regards

$\chi$ $\sigma$

 

[evinda]

please help!

(Wave)For your first question:We know that $\sum_{n=0}^{\infty} x^n= \frac{1}{1-x}, \text{ for } |x|<1$Replacing $x$ with $-3x$ we get:

$$\frac{1}{1-(-3x)}=\frac{1}{1+3x}= \sum_{n=0}^{\infty} (-3x)^n, \text{ for } |-3x|<1 \Rightarrow -1<3x<1 \Rightarrow \frac{-1}{3}<x< \frac{1}{3}$$
So what will be the value of $a_3$?

 

[Greg]

$$\frac{1}{1-(-3x)}=\frac{1}{1+3x}= \sum_{n=0}^{\infty} (-3x)^n, \text{ for } |-3x|<1 \Rightarrow -1<3x<1 \Rightarrow \frac{-1}{3}<x< \frac{1}{3}$$

Doesn't that imply \(\displaystyle 0^0=1\) (if \(\displaystyle x=0\))?

 

[evinda]

Doesn't that imply \(\displaystyle 0^0=1\) (if \(\displaystyle x=0\))?

$0^0 = 1$ is a convention that allows us to extend definitions in different areas of mathematics that otherwise would require treating $0$ as a special case.

 

[chisigma]

All the questions can be answered remembering the geometric series...

$\displaystyle \frac{1}{1 + \xi} = \sum_{n=0}^{\infty} (-1)^{n}\ \xi^{n},\ |\xi|< 1\ (1)$

Question 1: setting $\xi = 3\ x$ You have $\displaystyle \frac{2}{1 + 3\ x} = 2\ \sum_{n=0}^{\infty} (-1)^{n}\ (3\ x)^{n} \implies a_{3} = - 54$...

Question 2 : from (1) You derive...

$\displaystyle \frac{d}{d \xi} \frac{1}{1 + \xi} = - \frac{1}{(1 + \xi)^{2}} = \sum_{n=0}^{\infty} (-1)^{n}\ n\ x^{n-1}\ (2)$

... and setting $\xi = 4\ x$ You have $\displaystyle \frac{x}{(1 + 4\ x)^{2}}= \ \sum_{n=0}^{\infty} (-1)^{n}\ n\ 4^{n-1}\ x^{n} \implies a_{2} = 8$...

Question 3: from 1 You derive...

$\displaystyle \int_{0}^{\xi} \frac{d t}{1 + t} = \ln (1 + \xi) = \sum_{n=0}^{\infty} (-1)^{n} \frac{\xi^{n+1}}{n+1}\ (3)$

... so that You have $\displaystyle \int_{0}^{x} \xi^{2}\ \ln (1 + \xi)\ d \xi = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{n+4}}{(n+1)\ (n+4)} \implies a_{4} = \frac{1}{4}$

Kind regards

$\chi$ $\sigma$

 

[Nono713]

Alternatively, for the first problem, an "intuitive" - but not rigorous - derivation is as follows. Suppose that the following holds (given suitable radius of convergence etc etc):
$$\frac{2}{1 + 3x} = \sum_{n = 0}^\infty a_n x^n = a_0 + a_1 x + a_2 x^2 + \cdots$$
Then:
$$2 = (1 + 3x) \left ( a_0 + a_1 x + a_2 x^2 + \cdots \right )$$
Distribute to get:
$$2 = \left ( a_0 + a_1 x + a_2 x^2 + \cdots \right ) + 3x \left ( a_0 + a_1 x + a_2 x^2 + \cdots \right ) = \left ( a_0 + a_1 x + a_2 x^2 + \cdots \right ) + \left ( 3 a_0 x + 3 a_1 x^2 + 3 a_2 x^3 + \cdots \right )$$
Pairing up the coefficients of $x^n$ we get:
$$2 = a_0 + \left ( 3a_0 + a_1 \right ) x + \left ( 3a_1 + a_2 \right ) x^2 + \cdots$$
Or, more clearly:
$$2 + 0 x + 0 x^2 + \cdots = a_0 + \left ( 3a_0 + a_1 \right ) x + \left ( 3a_1 + a_2 \right ) x^2 + \cdots$$
Equate coefficients on both sides to find:
$$a_0 = 2$$
$$3a_0 + a_1 = 0 ~ ~ ~ \Rightarrow ~ ~ ~ a_1 = -3 a_0 = (-3) \times 2 = -6$$
$$3a_1 + a_2 = 0 ~ ~ ~ \Rightarrow ~ ~ ~ a_2 = -3 a_1 = (-3) \times (-6) = 18$$
$$3a_2 + a_3 = 0 ~ ~ ~ \Rightarrow ~ ~ ~ a_3 = -3 a_2 = (-3) \times 18 = -54$$
$$3a_3 + a_4 = 0 ~ ~ ~ \Rightarrow ~ ~ ~ a_4 = -3 a_3 = (-3) \times (-54) = 162$$
And so on. Eventually you find that $a_n = 2 \cdot (-3)^n$ as a general solution.

:D