Condensing infinite sums?

[Amaton]

Just wondering, but suppose [math]f(x)=\dfrac{a(x)}{b(x)}[/math].

 

If both [math]a(x)[/math] and [math]b(x)[/math] are each represented by their own infinite series, can I condense the function for both series into one sum and call it [math]f(x)[/math]?

[Bignose]

It really depends on the form of the sums. You might be able to, and you may not be able to. Do you have a specific example in mind?

[Amaton]

For example, [math]\tan\:x=\dfrac{\sin\:x}{\cos\:x}[/math]. Using this, I could show that [math]\tan\:x[/math] is equal to the quotient of the series representations for sine and cosine.

 

Say I have the power series for the sine divided by the power series for the cosine (an infinite sum over an infinite sum). Now can I condense this into a single sum and have the series functions in a (properly) combined quotient?

Edited October 13, 2012 by Amaton

[mathematic]

There is no easy way. If you look up the series for tan(x), you will the coefficients involve Bernoulli numbers.

 

http://en.wikipedia.org/wiki/Taylor_series

[Amaton]

There is no easy way. If you look up the series for tan(x), you will the coefficients involve Bernoulli numbers.

 

http://en.wikipedia....i/Taylor_series

 

Thanks. I see now.

 

I do notice the Bernoulli numbers, and something else...

What is [math]U_n[/math] supposed to denote?

[mathematic]

Thanks. I see now.

 

I do notice the Bernoulli numbers, and something else...

What is [math]U_n[/math] supposed to denote?

 

I didn't see [math]U_n[/math]. Where is it?

[Amaton]

Sorry for the confusion. It wasn't in the exact article you gave me. It was instead under the trigonometric functions Wiki, and for some reason I missed the denotation the first time reading it. [math]U_n[/math] denotes the "[math]n^{th}[/math] up/down number" as stated in the entry.

 

Anyway, here's a better formulation of my question.

 

Is it a valid to say that [math]f(x)=\dfrac{\displaystyle\sum_{k=a}^b g(x,k)}{\displaystyle\sum_{k=a}^b h(x,k)}[/math] is equal to [math]f(x)=\displaystyle\sum_{k=a}^b \frac{g(x,k)}{h(x,k)}[/math] in general?

[Bignose]

Is it a valid to say that [math]f(x)=\dfrac{\displaystyle\sum_{k=a}^b g(x,k)}{\displaystyle\sum_{k=a}^b h(x,k)}[/math] is equal to [math]f(x)=\displaystyle\sum_{k=a}^b \frac{g(x,k)}{h(x,k)}[/math] in general?

 

No

 

[math] \frac{\displaystyle\sum_{k=1}^2 x^k}{\displaystyle\sum_{k=1}^2 x^{2k}} = \frac{x+x^2}{x^2 + x^4} \ne \frac{x}{x^2} + \frac{x^2}{x^4} = \displaystyle\sum_{k=1}^2 \frac{x^k}{x^{2k}} [/math]

 

It is certainly not true in general, and as I wrote above, it really is only going to be special cases where you are going to be able to turn it into a single sum.

[Amaton]

Okay, thanks a lot