anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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\[\int\limits_{0}^{b}x*a^x da\]
anonymous
  • anonymous
I can't seem to do a u-sub that will work for me.
anonymous
  • anonymous
I want to assume that x is a constant and pull it outside, but I don't know if I can do that.

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anonymous
  • anonymous
you can, since youre differentiating with respect to a.
anonymous
  • anonymous
Because S = (0)a^(0) + (1)a^(1)...+(n)a^(n); you see why I think it's a constant?
anonymous
  • anonymous
Yes! But what happens to the other x?
anonymous
  • anonymous
wel, the other x is a constant too right?
anonymous
  • anonymous
whats the integral of a^x , if x is constant?
anonymous
  • anonymous
I have a wonderful solution to this, I believe.
anonymous
  • anonymous
Bring it :D
anonymous
  • anonymous
hm, you are right extremity
anonymous
  • anonymous
\[\int\limits_{0}^{b}a^x da\]?
anonymous
  • anonymous
\[ \sum_{x=0}^n xa^{\lambda x} = \sum_{x=0}^n \frac{d}{d\lambda} \frac{a^{\lambda x}}{\ln(a)} = \frac{1}{\ln(a)} \frac{d}{d\lambda} \sum_{x=0}^n a^{\lambda x}\] \[ = \frac{1}{\ln(a)} \frac{d}{d\lambda} \frac{1- a^{\lambda (n+1)}}{1- a^\lambda}\]
anonymous
  • anonymous
\[a^{x+1} /(x+1)\]
anonymous
  • anonymous
woah.
anonymous
  • anonymous
where'd you get lambda?
anonymous
  • anonymous
just made it up. Set it to 1 when you're all done.
anonymous
  • anonymous
it's to differentiate x from the exponent?
anonymous
  • 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.
anonymous
  • anonymous
Oops, I did make a typo
anonymous
  • anonymous
should be \[ \frac{na^{n+2} - na^{n+1} - a^{n+1} + a}{(1-a)^2} \]
anonymous
  • anonymous
And wolfram alpha confirms. Awesome.
anonymous
  • anonymous
By the way, if a = 1, obviously the sum is just \[ \sum_{x=0}^n x = \frac{n(n+1)}{2}\]
anonymous
  • anonymous
lol
anonymous
  • anonymous
Thanks! I am going back over this to make sure I understand. Thanks for all your help!
anonymous
  • anonymous
No problem. Be careful with your derivatives. And @TuringTest / @Gravion should recognize this technique.... ;)
anonymous
  • anonymous
:D
anonymous
  • anonymous
Are you still on?
anonymous
  • 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
anonymous
  • anonymous
the integral doesn't converge, so...
anonymous
  • anonymous
That is the hint. I integrated, then differentiated.
anonymous
  • anonymous
And it should converge if a is sufficiently small.
anonymous
  • anonymous
hm...
anonymous
  • anonymous
If |a| < 1, that reduces to \[ \frac{a}{(1-a)^2} \]
anonymous
  • anonymous
I'm sorry, I'm thrown off by the ln(a). I can't get it in my own work...
anonymous
  • 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!
anonymous
  • anonymous
Did you follow my reasoning up until the part where you take the derivative?
anonymous
  • anonymous
um, actually no. I don't understand d/dlambda
anonymous
  • 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?
anonymous
  • 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!
anonymous
  • 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.
anonymous
  • anonymous
It's not a trivial series to sum, what class are you doing this for?

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