- #1

- 152

- 0

1/(1+x^2)=1-x^2 + x^4-x^6+....+(-1)^n(x^2n)+.... -1 > x > 1

how to get this??.....

how to get this??.....

- Thread starter newton1
- Start date

- #1

- 152

- 0

1/(1+x^2)=1-x^2 + x^4-x^6+....+(-1)^n(x^2n)+.... -1 > x > 1

how to get this??.....

how to get this??.....

- #2

- 1,561

- 3

f(x) = f(0) + x*f'(0) + (x^2)/2! * f''(0) +....

Which derives the binomial series:

(1+a)^n = 1+ na + n(n-1)/2! * a^2 + n(n-1)(n-2)/3! * a^3 .... As long as |a|<1

Substitute a = x^2 and n = -1 et viola!

- #3

dg

If you do not want to go all the way to the general theory of series of powers (Taylor and MacLaurin series), you can simply use the result of the geometric seriesOriginally posted by newton1

1/(1+x^2)=1-x^2 + x^4-x^6+....+(-1)^n(x^2n)+.... -1 > x > 1

how to get this??.....

1/(1-y)=1+y+y^2+...+y^n+...

which can be obtained with basic arithmetic arguments and substistute y with -x^2.

- #4

- 508

- 0

Yes. Because, if you abbreviate the right hand side as S, then obviously

y*S = S-1

y*S = S-1

- Last Post

- Replies
- 3

- Views
- 6K

- Last Post

- Replies
- 1

- Views
- 3K

- Replies
- 3

- Views
- 2K

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 1K

- Replies
- 6

- Views
- 575