- #1

monsmatglad

- 76

- 0

please help

I have tried this several times, but it doesn't match with the true answer.

thank you

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In summary, the series for u^(n-1) is\sum_{n=1}^\infty \frac{u^n}{n}, which is easily recognizable as the integral of u^(n-1). Now, do you know the series for u^(n-1)?

- #1

monsmatglad

- 76

- 0

please help

I have tried this several times, but it doesn't match with the true answer.

thank you

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- #2

Char. Limit

Gold Member

- 1,222

- 22

Try substituting in u=2x - 1 for now, to get [itex]\sum_{n=1}^\infty \frac{u^n}{n}[/itex], which is easily recognizable as the integral of u^(n-1). Now, do you know the series for u^(n-1)?

- #3

monsmatglad

- 76

- 0

what I tried was to derivate to get rid of the n under, which then gave me 2*(2x-1)^n-1. Then I used the sum of a geometric series, getting: 2/(1-(2x-1))= 2/(2-2x)=1/(1-x).

Then i integrated back 1/(1-t) with t going from 0 to x. Then I get -ln(1-t), but this does not fit with what the answer should be.

- #4

Char. Limit

Gold Member

- 1,222

- 22

Actually, the solution IS correct, up to a constant. The problem is, by deriving, you lost the constant. So all you need to do now is figure out what that constant term is.

- #5

monsmatglad

- 76

- 0

that's good news:P but I don't see how to find the constant term. The true answer should be -ln(2-2x) .

- #6

Char. Limit

Gold Member

- 1,222

- 22

Try factoring that. You get -ln(2 (1 - x)). Using the rules of logarithms, how can you factor that out?

- #7

monsmatglad

- 76

- 0

-(ln2+ln(1-x)), but i don't see how to derive this without going through the true answer.

- #8

monsmatglad

- 76

- 0

I mean, how do I get to the constant without using the correct answer to derive from.

- #9

monsmatglad

- 76

- 0

That might have been a bit fuzzy due to some sloppiness with names, but basically I need to know how to find that constant term that disappears when deriving.

- #10

- 22,183

- 3,322

So you've figured out correctly that

[tex]\sum_{n=1}^{+\infty}{\frac{(2x-1)^n}{n}}=-ln(1-x)+C[/tex]

So, if you fill in a particular x, then this equality must hold. You might fill in 0 for x, but then you'll have to evaluate the series

[tex]\sum_{n=1}^{+\infty}{\frac{(-1)^n}{n}}[/tex]

which may be a little difficult. Can you think of an x that will make the serie easy? (for example, an x such that all the terms of the series are 0?)

- #11

monsmatglad

- 76

- 0

x=2?

- #12

monsmatglad

- 76

- 0

ehm. x=2 can't be correct, but with x=0, you get ln(1)=0

right?

- #13

monsmatglad

- 76

- 0

-1/2?

sorry, I'm stupid

- #14

monsmatglad

- 76

- 0

that was supposed to be 1/2 without the minus

- #15

- 22,183

- 3,322

Will the series be easy if you fill in x=2? You will get the series

[tex]\sum_{n=1}^{+\infty}{\frac{3^n}{n}}[/tex]

which isn't easy to determine.

In fact, you probably have somewhere in your assignment the contraint that [itex]0<x<1[/itex]. So you can't fill in 2 since the series will diverge for that value!

- #16

- 22,183

- 3,322

monsmatglad said:that was supposed to be 1/2 without the minus

Yes! That is good. What do you get for x=1/2?

- #17

monsmatglad

- 76

- 0

ah, so any value that makes the serie converge will do?

- #18

- 22,183

- 3,322

monsmatglad said:ah, so any value that makes the serie converge will do?

Well, yes and no. Every value that makes the series converge will do in the sense that it gives you a value for C.

However, you will want to pick a value for x that makes the serie easy. If you pick x=1/2, then this is the case since the series will vanish. So this is the ideal value to take! You can pick x=3/4 if you want, but the serie will be much more difficult to evaluate!

- #19

monsmatglad

- 76

- 0

so, i pick 1/2. How do I get from that to getting the constant -ln2?

- #20

- 22,183

- 3,322

Well, substitute 1/2 in your equation. What do you get?

- #21

Char. Limit

Gold Member

- 1,222

- 22

Remember first that you have f(x) = -ln(1 - x) + C, and you want to find C.

- #22

monsmatglad

- 76

- 0

-ln(1/2)=ln2? that's right isn't it?

- #23

monsmatglad

- 76

- 0

i think i have got it. i get 0 on one side av ln2 + C on the other. moving them around gives - ln2. correct?

- #24

Char. Limit

Gold Member

- 1,222

- 22

That's correct. So now you have your constant term, and can write the answer to the series.

- #25

monsmatglad

- 76

- 0

thank you very much for being so patient.=)

This phrase refers to the sum of all the terms in a mathematical series, where the starting value of the series is n=1 and the series continues infinitely.

Infinite series can be difficult to solve because they involve an infinite number of terms, which makes it challenging to find a specific value for the sum. This requires the use of advanced mathematical techniques such as convergence tests and summation formulas.

The sum of an infinite series can be found by using various methods such as the geometric series formula, telescoping series, or the ratio test. These methods involve evaluating the series term by term and determining if the series converges or diverges.

Finding the sum of an infinite series is important in various mathematical applications, such as calculating probabilities, determining the convergence of numerical methods, and solving differential equations. It also helps in understanding the behavior of infinite sequences and series.

Yes, an infinite series can have a finite sum if the series is convergent. This means that the terms in the series approach a finite value as the number of terms increases, and the sum of the series can be calculated precisely. However, if the series is divergent, the sum will be infinite or undefined.

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