Amaton Posted October 13, 2012 Share Posted October 13, 2012 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]? Link to comment Share on other sites More sharing options...
Bignose Posted October 13, 2012 Share Posted October 13, 2012 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? Link to comment Share on other sites More sharing options...
Amaton Posted October 13, 2012 Author Share Posted October 13, 2012 (edited) 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 Link to comment Share on other sites More sharing options...
mathematic Posted October 13, 2012 Share Posted October 13, 2012 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 Link to comment Share on other sites More sharing options...
Amaton Posted October 14, 2012 Author Share Posted October 14, 2012 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? Link to comment Share on other sites More sharing options...
mathematic Posted October 14, 2012 Share Posted October 14, 2012 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? Link to comment Share on other sites More sharing options...
Amaton Posted October 16, 2012 Author Share Posted October 16, 2012 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? Link to comment Share on other sites More sharing options...
Bignose Posted October 16, 2012 Share Posted October 16, 2012 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. 2 Link to comment Share on other sites More sharing options...
Amaton Posted October 24, 2012 Author Share Posted October 24, 2012 Okay, thanks a lot Link to comment Share on other sites More sharing options...
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