## 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

1. anonymous

The 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$\int\limits_{0}^{b}x*a^x da$

2. anonymous

I can't seem to do a u-sub that will work for me.

3. anonymous

I want to assume that x is a constant and pull it outside, but I don't know if I can do that.

4. anonymous

you can, since youre differentiating with respect to a.

5. anonymous

Because S = (0)a^(0) + (1)a^(1)...+(n)a^(n); you see why I think it's a constant?

6. anonymous

Yes! But what happens to the other x?

7. anonymous

wel, the other x is a constant too right?

8. anonymous

whats the integral of a^x , if x is constant?

9. anonymous

I have a wonderful solution to this, I believe.

10. anonymous

Bring it :D

11. anonymous

hm, you are right extremity

12. anonymous

$\int\limits_{0}^{b}a^x da$?

13. anonymous

$\sum_{x=0}^n xa^{\lambda x} = \sum_{x=0}^n \frac{d}{d\lambda} \frac{a^{\lambda x}}{\ln(a)} = \frac{1}{\ln(a)} \frac{d}{d\lambda} \sum_{x=0}^n a^{\lambda x}$ $= \frac{1}{\ln(a)} \frac{d}{d\lambda} \frac{1- a^{\lambda (n+1)}}{1- a^\lambda}$

14. anonymous

$a^{x+1} /(x+1)$

15. anonymous

woah.

16. anonymous

where'd you get lambda?

17. anonymous

just made it up. Set it to 1 when you're all done.

18. anonymous

it's to differentiate x from the exponent?

19. anonymous

yep. I may have made a typo, but I'm getting $\frac{a(n+1) -a^n - n}{(1-a)^2}$ as a final result. Assuming of course that a does not equal 1.

20. anonymous

Oops, I did make a typo

21. anonymous

should be $\frac{na^{n+2} - na^{n+1} - a^{n+1} + a}{(1-a)^2}$

22. anonymous

And wolfram alpha confirms. Awesome.

23. anonymous

By the way, if a = 1, obviously the sum is just $\sum_{x=0}^n x = \frac{n(n+1)}{2}$

24. anonymous

lol

25. anonymous

Thanks! I am going back over this to make sure I understand. Thanks for all your help!

26. anonymous

No problem. Be careful with your derivatives. And @TuringTest / @Gravion should recognize this technique.... ;)

27. anonymous

:D

28. anonymous

Are you still on?

29. anonymous

I 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

30. anonymous

the integral doesn't converge, so...

31. anonymous

That is the hint. I integrated, then differentiated.

32. anonymous

And it should converge if a is sufficiently small.

33. anonymous

hm...

34. anonymous

If |a| < 1, that reduces to $\frac{a}{(1-a)^2}$

35. anonymous

I'm sorry, I'm thrown off by the ln(a). I can't get it in my own work...

36. anonymous

I'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!

37. anonymous

Did you follow my reasoning up until the part where you take the derivative?

38. anonymous

um, actually no. I don't understand d/dlambda

39. anonymous

you said you made lamda up---to differentiate from the other x (I guess) but then how do you also do a derivative for it?

40. anonymous

wait, 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!

41. anonymous

I 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.

42. anonymous

It's not a trivial series to sum, what class are you doing this for?