Math Induction

[Chiraq]

Can you help me to solve this
$\displaystyle \sum_{j=1}^n (tg(x/2))/2 = ctg (x/2^n)/2^n -ctg x$
Thanks in advance

 

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[topsquark]

Can you help me to solve this
\sum (from j=1 to n) (tg(x/2))/2 = ctg (x/2^n)/2^n -ctg x
Thanks in advance

What you wrote is this:
$$\sum_{j = 1}^n \dfrac{tg \left ( \dfrac{x}{2} \right )}{2} = \dfrac{ctg \left ( \dfrac{x}{2^n} \right )}{2^n} - ctg(x)$$
There is no "j" in the summand. Please fix the problem statement.

-Dan

 

[Chiraq]

What you wrote is this:
$$\sum_{j = 1}^n \dfrac{tg \left ( \dfrac{x}{2} \right )}{2} = \dfrac{ctg \left ( \dfrac{x}{2^n} \right )}{2^n} - ctg(x)$$
There is no "j" in the summand. Please fix the problem statement.

-Dan

This is setup, I'm not sure if this is mistake. I was thinking maybe there is mistake, but this is setup in my book. It's maybe tg (x/2)/2^j

 

[Steven G]

If you set is correct, then the left hand side equals n((tg(x/2))/2)

$\displaystyle \sum_{i=1}^n a = a + a + a + ... + a = na$.
Your a is simply (tg(x/2))/2

 

[Deleted member 4993]

This is setup, I'm not sure if this is mistake. I was thinking maybe there is mistake, but this is setup in my book. It's maybe tg (x/2)/2^j

As posted the problem does not make sense. Please post a picture of your assignment/problem as it was given you.

Why does title of your post include the term 'induction '?

 

[Chiraq]

If you set is correct, then the left hand side equals n((tg(x/2))/2)

$\displaystyle \sum_{i=1}^n a = a + a + a + ... + a = na$.
Your a is simply (tg(x/2))/2

I know, I tried to solve it that way, but couldn't

 

[topsquark]

I know, I tried to solve it that way, but couldn't

Seeing as the original statement is incorrect for any value of n this does not surprise me. You will have to find a corrected version of the problem before you can do it.

-Dan