Hard CALCULUS 3 Problem

[LightRay7]

Hello,
I would appreciate an early reply.


Consider the expression given at http://postimage.org/image/5g5r4uvhb/ .


1) Expand the expression as a series and show that the coefficient of x^(k+1) is 1^k + 2^k + 3^k + ... + (n-1)^k. (Hint: expand every e^(rx) as a series and gather like terms.)
2) Use geometric series to show that the expression can also be written as [x/(e^x -1) ]k!(e^(nx) - 1)
3) Expand both parts of the expression in 2) as series (writing the generating function in terms of the B<SUB>n, take the Cauchy product to find the coefficient of x^(k-1). Check that your answer gives the right formula.
4) Bernoulli calculated B<SUB>0 to B<SUB>11 in order to use his formula to show that:
1^10 + 2^10 + 3^10 + ... + 1000^10 = 91409924241424243424241924242500
He claimed that it took him 7.5 minutes to do this by hand. Verify that his answer is correct.

Here's the work I did.
The expression is:

k!x * "sigma"e^(rx) from r=0 to (n-1). Basically it's k!x multiplied by the sum of e^(rx) from r=0 to (n-1)

Given that e^x = 1 + x + x^2/2! + x^3/3! + ...
Expanding: k!x(e^x + e^(2x) + e^(3x) + ...+ e^((n-1)x))

= k!x((1+x^2/2! + x^3/3! + ...) + (1+ 2x + 2x^2 + 4x^3/3+ ...) + ...+ (1 + (n-1)x + [(n-1)^2*x^2]/2 +...)

okay what do I do from here, how do I gather like terms. How does x^k+1 pop out? T hank you

 

[Deleted member 4993]

Hello,
I would appreciate an early reply.


Consider the expression given at http://postimage.org/image/5g5r4uvhb/ .


1) Expand the expression as a series and show that the coefficient of x^(k+1) is 1^k + 2^k + 3^k + ... + (n-1)^k. (Hint: expand every e^(rx) as a series and gather like terms.)
2) Use geometric series to show that the expression can also be written as [x/(e^x -1) ]k!(e^(nx) - 1)
3) Expand both parts of the expression in 2) as series (writing the generating function in terms of the B<SUB>n, take the Cauchy product to find the coefficient of x^(k-1). Check that your answer gives the right formula.
4) Bernoulli calculated B<SUB>0 to B<SUB>11 in order to use his formula to show that:
1^10 + 2^10 + 3^10 + ... + 1000^10 = 91409924241424243424241924242500
He claimed that it took him 7.5 minutes to do this by hand. Verify that his answer is correct.

We do not open unknown attachments.

In addition:

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

Duplicate Post:

http://answers.yahoo.com/question/index?qid=20120509063010AA5I5wH

http://mathhomeworkanswers.org/26122/calculus-3-series

http://mathhelpforum.com/calculus/198581-calculus-3-problem-series.html

 

[LightRay7]

This is not an attachment. I could not write the expression in Text form. Therefore, it is a link to an image-form of the expression.
I admit I posted this question at other places. This is because I am in a rush and need to get help as soon as possible for these problems.

Thank you for your reply

Please write back if you know how to solve these problems. I haven't got a clue on how to do them.

We do not open unknown attachments.

In addition:

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

Duplicate Post:

http://answers.yahoo.com/question/index?qid=20120509063010AA5I5wH

http://mathhomeworkanswers.org/26122/calculus-3-series

http://mathhelpforum.com/calculus/198581-calculus-3-problem-series.html

 

[Deleted member 4993]

Expand the expression as a series and show that the coefficient of x^(k+1) is 1^k + 2^k + 3^k + ... + (n-1)^k. (Hint: expand every e^(rx) as a series and gather like terms.)

This is straight-forward and explicit instruction. Did you expand e^rx ?

What have you done with it? - share your work.

 

[mmm4444bot]

Please write back if you know how to solve these problems.

Well, of course there are people here who know how to solve these problems. :???:

Is camera-ready copy the type of "help" that you expect to receive here? If so, you're at the wrong web site. We tutor here.

If you've been misled by other contributors on these boards who post completed exercises for people needing help, then you should know that those users are violating our forum rules.

We do not do homework.

 

[LightRay7]

my work

Well, of course there are people here who know how to solve these problems. :???:

Is camera-ready copy the type of "help" that you expect to receive here? If so, you're at the wrong web site. We tutor here.

If you've been misled by other contributors on these boards who post completed exercises for people needing help, then you should know that those users are violating our forum rules.

We do not do homework.

Here's the work I did.
The expression is:

k!x * "sigma"e^(rx) from r=0 to (n-1). Basically it's k!x multiplied by the sum of e^(rx) from r=0 to (n-1)

Given that e^x = 1 + x + x^2/2! + x^3/3! + ...
Expanding: k!x(e^x + e^(2x) + e^(3x) + ...+ e^((n-1)x))

= k!x((1+x^2/2! + x^3/3! + ...) + (1+ 2x + 2x^2 + 4x^3/3+ ...) + ...+ (1 + (n-1)x + [(n-1)^2*x^2]/2 +...)

okay what do I do from here, how do I gather like terms. How does x^k+1 pop out? T hank you