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anonymous
 4 years ago
If I need to find a closed solution for the summation of x*a^x from x = 0 to n, what do I do? For starters, I've decided to do an integral from 0 to b of the same function, but now I have 2 variables! ah! Help please :D
anonymous
 4 years ago
If I need to find a closed solution for the summation of x*a^x from x = 0 to n, what do I do? For starters, I've decided to do an integral from 0 to b of the same function, but now I have 2 variables! ah! Help please :D

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The only hint I was given was to take the integral of the function I was trying to take the summation of. But, now I have\[\int\limits_{0}^{b}x*a^x da\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can't seem to do a usub that will work for me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I want to assume that x is a constant and pull it outside, but I don't know if I can do that.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can, since youre differentiating with respect to a.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because S = (0)a^(0) + (1)a^(1)...+(n)a^(n); you see why I think it's a constant?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes! But what happens to the other x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wel, the other x is a constant too right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whats the integral of a^x , if x is constant?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have a wonderful solution to this, I believe.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hm, you are right extremity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{b}a^x da\]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \sum_{x=0}^n xa^{\lambda x} = \sum_{x=0}^n \frac{d}{d\lambda} \frac{a^{\lambda x}}{\ln(a)} = \frac{1}{\ln(a)} \frac{d}{d\lambda} \sum_{x=0}^n a^{\lambda x}\] \[ = \frac{1}{\ln(a)} \frac{d}{d\lambda} \frac{1 a^{\lambda (n+1)}}{1 a^\lambda}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where'd you get lambda?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just made it up. Set it to 1 when you're all done.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's to differentiate x from the exponent?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep. I may have made a typo, but I'm getting \[ \frac{a(n+1) a^n  n}{(1a)^2} \] as a final result. Assuming of course that a does not equal 1.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oops, I did make a typo

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0should be \[ \frac{na^{n+2}  na^{n+1}  a^{n+1} + a}{(1a)^2} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And wolfram alpha confirms. Awesome.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0By the way, if a = 1, obviously the sum is just \[ \sum_{x=0}^n x = \frac{n(n+1)}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks! I am going back over this to make sure I understand. Thanks for all your help!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No problem. Be careful with your derivatives. And @TuringTest / @Gravion should recognize this technique.... ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have a question: what if I need to use the hint (i.e., that's what they want us to do). I kinda see what you are doing, but I can move on from the integral from 0 to infinity of x*a^x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the integral doesn't converge, so...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That is the hint. I integrated, then differentiated.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And it should converge if a is sufficiently small.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If a < 1, that reduces to \[ \frac{a}{(1a)^2} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, I'm thrown off by the ln(a). I can't get it in my own work...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm getting a^(x+1)/ (x+1) though, which I can use on the limits of intergration...but then I get x[infinity  a] which doesn't make sense!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Did you follow my reasoning up until the part where you take the derivative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0um, actually no. I don't understand d/dlambda

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you said you made lamda upto differentiate from the other x (I guess) but then how do you also do a derivative for it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait, the "integrate then differentiate" thing is what I am supposed to do (that's the hint they gave" but I dunno why I can't understand this!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I invented a parameter to put in the problem, differentiated and integrated with respect to said parameter, and at the end of the day, after everything was done, I set that parameter to 1.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's not a trivial series to sum, what class are you doing this for?
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