Question

questions from calculus, anybody ?

Answer 1

Can i post the qns @hartnn

Answer 2

sure

Answer 3

only one at a time would be great :)

Answer 4

sorry Hw 11 please

Answer 5

qns 2a and d

Answer 6

question 2a and d please.

Answer 7

for 2a :

y = 1 + x + 5x^2 + 13x^3 + ...
-2xy = -2x - 2x^2 - 10x^3 - ...
-3x^2y = - 3x^2 - 3x^3 - ...
-----------------------------------

Answer 8

add them up vertically, what do u get ?

Answer 9

-3x^2-2x+1

Answer 10

1-x over the above ? @hartnn

Answer 11

@ganeshie8

Answer 12

y = 1 + x `+ 5x^2` + 13x^3 + ...
-2xy = -2x `- 2x^2` - 10x^3 - ...
-3x^2y = `- 3x^2` - 3x^3 - ...
-----------------------------------
`0`

Answer 13

do you see that adding up that column gives you 0 ?

Answer 14

similarly adding ANY column to the right of that column gives you 0

Answer 15

y = 1 + x `+ 5x^2` `+ 13x^3` + ...
-2xy = -2x `- 2x^2` `- 10x^3` - ...
-3x^2y = `- 3x^2` `- 3x^3` - ...
-----------------------------------
1-x + `0` + `0` + ...

Answer 16

Thank you and taking y as common and we get the proof that y = 1- x / -3x^2-2x + 1

Answer 17

?

Answer 18

yes

Answer 19

thanks a lot sir.

Answer 20

And for qns 2d?

Answer 21

are you done with b and c ?

Answer 22

yes am good with b and c

Answer 23

\[\large \rm y = \sum\limits_{n=0}^{\infty }a_nx^n\]

Notice that your actual goal here is to find the explicit formula for the coefficient generating function : \(a_n\)

Answer 24

\[\large \rm y = \sum\limits_{n=0}^{\infty }a_nx^n = \dfrac{A}{1-3x} + \dfrac{B}{1+x}\]

Answer 25

what are the values of A and B ?

Answer 26

A and B both are 1/2

Answer 27

very good, next use the hint from part c

Answer 28

\[\large \rm y = \sum\limits_{n=0}^{\infty }a_nx^n = \frac{1}{2}\left(\dfrac{1}{1-3x} + \dfrac{1}{1+x} \right) \]

Answer 29

for part c i wrote the infinite series expansion for 1 / 1-X which is 1 + x + x^ 2 + x^3 and replaced x with negative x to find 1/ 1 +x

Answer 30

you need to add up the terms in both series and compare to find the formula for coefficients

Answer 31

\[\rm \begin{align}y = \sum\limits_{n=0}^{\infty }a_nx^n &= \frac{1}{2}\left[\dfrac{1}{1-3x} + \dfrac{1}{1+x} \right] \\~\\
&= \frac{1}{2}\left[1+3x+(3x)^2 + \cdots + 1-x+x^2-x^3 + \cdots \right]\\~\\
&= \sum\limits_{n=0}^{\infty}\frac{1}{2}\left[ (3x)^n + (-x)^n \right] \\~\\
&= \sum\limits_{n=0}^{\infty}\frac{1}{2}\left[ 3^n + (-1)^n \right]x^n \\~\\

\end{align}\]

Answer 32

compare

Answer 33

Thank you so much.

Answer 34

np :)

Answer 35

I had to submit the unfinished ones today, and thanks for helping me out.

Answer 36

Try this if you have time :

find an explicit formula using the exact same trick as above
\[\rm a_0=0, a_1 = 1, \\a_n = a_{n-1} + a_{n-2}\]

Answer 37

let me figure it out, and is there any chance of helping me with the extra credit question?

Answer 38

The question 3 ?