Harder 3U: Logs & the Mercator Series (1 Viewer)

[KFunk]

Hey guys, this is a question I cam up against in the extension section of the cambridge 3U book that I borrowed recently. I'll try and work it up to where I got to and let you guys have at it from there.

1) First you have to prove that:

1 - t + t² - t³ + ... + t²ⁿ = 1/(1+t) + t²ⁿ⁺¹/(1+t), for t≠1

you can do that easily by using the the sum of a GP, so not much of a problem

2) Integrate both sides of the result from t=0 to t=x to show that for x>-1 :

log_e(1+x) = x - x²/2 + x³/3 + x⁴/4 + ... + x²ⁿ⁺¹/(2n+1) - ∫ t²ⁿ⁺¹/(1+t) dt
where the integral is between 0 and x ... this isn't very hard to show but from that --->

3) Explain why t²ⁿ⁺¹/(1+t) ≤ t²ⁿ⁺¹, for o≤t≤1. Hence prove that for 0≤x≤1, the integral ∫ t²ⁿ⁺¹/(1+t)dt [between 0 and x] converges to 0 as n --> ∞ . Hence show that

log(1+x) = x - x²/2 + x³/3 + x⁴/4 + ... + x²ⁿ⁺¹/(2n+1)

So good luck doing the question. I suspect there must be another aproach to the one I took because I ended up, after evaluating that integral, with the expression I am trying to prove and you can't really assume the end of the proof as part of it. Essentially it's showing that:

{from r=1 --> ∞} ∑ (-1)^r+1/r.x^r = ln(1+x)

Any input is appreciated, thanks in advance.

 

[shafqat]

umm i have a worked solution to this question, but ill let others try first

 

[Slidey]

Explain why t^(2n+1)/(1+t) ≤ t^(2n+1), for o≤t≤1. Hence prove that for 0≤x≤1, the integral ∫ t^(2n+1)/(1+t)dt [between 0 and x] converges to 0 as n --> ∞ .

Since ∫g(x)dx=0 iff g(x)dx=0, lim(t^(2n+1)/(1+t)) as n->inf must equal zero.
t^(2n+1)/(1+t) <= t^(2n+1)
lim(t^(2n+1)) as n-> inf = 0 (t is less than 1),
.'. lim(t^(2n+1)/(1+t)) as n->inf = 0
So ∫ x^(2n+1)/(1+x)dt = 0, as n-> infinity

Except for the case t=1, which does not work, and was previously excluded, yet which the current part of the question includes...

 

[KFunk]

umm i have a worked solution to this question, but ill let others try first

Would you be able to hint me in the right direction? (and I'm talking a fairly direct, not so subtle hint )

 

[martin]

3) Explain why t^(2n+1)/(1+t) ≤ t^(2n+1), for 0≤t≤1. Hence prove that for 0≤x≤1, the integral ∫ t2n+1/(1+t)dt [between 0 and x] converges to 0 as n --> ∞ .

0≤t≤1
1≤t+1≤2
1≥1/(t+1)≥1/2

so 1/(t+1)≤1 therefore t^(2n+1)/(t+1) ≤ t^(2n+1)

therefore integral from 0 to x of LHS ≤ integral of RHS = x^(2n+2)/(2n+2)
and x^(2n+2)/(2n+2) ≤ x^(2n+2) since n is positive

and x^(2n+2) -> 0 as n->∞ because 0≤x≤1

since the integral we are interested in is less than or equal to something that goes to zero the integral goes to zero also.

So going to infinity we get

log(1+x) = x - x^2/2 + x^3/3 - .... (note that this is an infinite sum and it converges for |x|<1)

 

[Slidey]

and x^(2n+2) -> 0 as n->∞ because 0≤x≤1

since the integral we are interested in is less than or equal to something that goes to zero the integral goes to zero also.

Suppose x=1.

 

[martin]

Yeah, you're right Slide Rule, I didn't think about x=1. My answer doesn't cover it and I'm not sure how you could show that the integral goes to 0. Edit:see next message

However maybe we don't need to show that. At x = 1 you get

log2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ....

Put 10 terms or so into a calculator and you should see it gets reasonably close. To prove that it converges (this is first or second year uni stuff) you just use the alternating series test: If a_n -> 0 as n->∞ then the sum from i=1 to ∞ of (-1)^n a_n converges.

The basic idea of the proof is that the partial sums oscillate around the final sum. 1 - 1/2 + 1/3 = 0.833 and 1 - 1/2 + 1/3 - 1/4 = 0.583. log2=0.693 is in between these two and since the numbers being added and subtracted are getting smaller it gets closer and closer.

As a sidenote 1 + 1/2 + 1/3 + 1/4 + ... doesn't converge, it goes to infinity very slowly. This means that 1 - 1/2 + 1/3 - .... has different values if you rearrange it (look up conditionally convergent series if you're interested).

 

[martin]

Actually after reading over it again its pretty easy.

from above,
the integral ≤ x^(2n+2)/(2n+2)

if x=1, x^(2n+2)/(2n+2) = 1/(2n+2) -> 0 as n->∞

 

[KFunk]

Haha, well done. That sure makes things a lot simpler.

 

[Slidey]

Haha. Nice, martin.