1. Originally Posted by mkeg1
I am confused because the original problem gives the integral with n=0,1,2,.... then it asks for a recurrence formula for y sub k to y sub k-1 for k 1,2,3,.... am I still plugging in the same number?
You are meant to get an expression for $y_n$ in terms of $y_{n-1}$ and n. Please go back and carefully review the thread. Everything is there, all you have to do is put together.

2. Originally Posted by mr fantastic
You are meant to get an expression for $y_n$ in terms of $y_n$ and n. Please go back and carefully review the thread. Everything is there, all you have to do is put together.
$y_n$ should be in terms of $y_{n-1}$ and $n$

3. $y_{n-1}=-t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}$ for n=1,2,3....

?

4. Remember that $y_n = \int{e^{-t}t^{n}\,dt}$.

You have shown $\int{e^{-t}t^n\,dt} = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}$.

So $y_n = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}$.

Can you see that $\int{e^{-t}t^{n-1}\,dt}$ is almost the same as your original integral - the only difference being that $n$ is now $n-1$.

Surely this is $y_{n-1}$.

So what do you think $y_n$ is in terms of $y_{n-1}$?

5. $y_{n-1}=-t^{n-1}e^{-t} + (n-1)\int{e^{-t}t^{n-2}\,dt}$

6. Originally Posted by mkeg1
$y_{n-1}=-t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}$ for n=1,2,3....

?
Did you read post #12 at all? A lot of time is starting to be spent saying the same things. Please .... take some time to think about the probelm and what has been posted in this thread.

7. Yes, I read post 12, I am not very good at math as you can see. Maybe I just need to spend a little more time evaluating the problem

8. Originally Posted by mkeg1
$y_{n-1}=-t^{n-1}e^{-t} + (n-1)\int{e^{-t}t^{n-2}\,dt}$
I suggest you read post number 19...

9. Originally Posted by Prove It
I suggest you read post number 19...
$y_n$ is a definite integral not an indefinite integral.

10. $y_n = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}$

Kind of a shot in the dark because I sill dont understand, but would i replace the integral part with y sub n-1?

11. Originally Posted by mkeg1
since $dv=e^{-t}dt$ i took the integral and got $-(1/e^{t}loge)$

I know integral of e^x is e^x but doesn tthe negative sign change it?
That "log" is the natural logarithm and log(e)= 1!

I suspect you looked up the integral of $a^x$. You should know that $\int e^x dx= e^x+ C$ without having to look it up. If you let u= -x, then du= -dx so that $\int e^{-x}dx= -\int e^u du= -e^u+ C= -e^{x}+ C$.

12. Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,

... is lazy integration by parts, doing without u and v.

Now, of course, state $y_{k}$ in terms of $y_{k-1}$ and k.
_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

13. Hey guys I really appreciate all the help.

I understand the integration, I just dont understand what to do when applying $y_n$in terms of $y_{n-1}$

14. Can you at least see the pattern in

$y_{1} = y_{0} - e^{-1}$

$y_{2} = 2y_{1} - e^{-1}$

$y_{3} = 3y_{2} - e^{-1}$

and write it as

$y_{k} =$ ... ?

(or $y_{n} =$ ... ?)

I don't think any more is required than that. (And the starting value, for which, see above.)

15. Originally Posted by mkeg1
$y_n = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}$

Kind of a shot in the dark because I sill dont understand, but would i replace the integral part with y sub n-1?
No. For what seems like the one millionth time: $y_n$ is a DEFINITE integral!

$\displaystyle y_n = \left[-t^ne^{-t}\right]_{0}^{1} + n \int_0^1 {e^{-t}t^{n-1}\,dt}$.

Now the integral on the right hand side can be replaced with $y_{n-1}$. The first term on the right hand side needs to be evaluated in the usual way.

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