- #1

- 191

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Any ideas ?

- Thread starter bogdan
- Start date

- #1

- 191

- 0

Any ideas ?

- #2

- 166

- 1

The sum from k to infinity of (x/(n*(n+1))) is x/k, so you could sum up to a certain point and then use this to approximate the truncation error.

- #3

- 34

- 1

I think the answer is tan(x).

sin(x/(n(n+1))) = sin(x/n)cos(x/(n+1)) - sin(x/(n+1))cos(x/n)

After some simplifications you get:

sum tan(x/n)-tan(x/(n+1))

That is:

tan x - tan x/2 +

+ tan x/2 - tan x/3

...

which is

tan x - tan 0 = tan x

- #4

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(eram doar curios sa vad cine stie sa-l rezolve...e dintr-o carte de exercitii de analiza...cu tot cu raspunsuri)

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