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The Maclaurin Series...

Let f(x)=ln⁡(1+x)f(x)=\ln(1+x), x>−1x>-1.

Use the Lagrange form of the remainder to prove that ff is represented by its Maclaurin series in the interval [0,1][0,1].


I said: If ff is represented by its Maclaurin series in the interval [0,1][0,1] then for x∈[0,1]x\in[0,1],

Unparseable latex formula:

\diplaystyle \lim_{n\to\infty} R_{n,0}f(x)=0

.

The Lagrange form of the remainder,

Rn,0f(x)=f(n+1)(c)(n+1)!(x−0)n+1=(−1)n+2(1+c)n+1(n+1)!xn+1R_{n,0}f(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-0)^{n+1}=\frac{(-1)^{n+2}}{(1+c)^{n+1}(n+1)!}x^{n+1} for some c∈(0,x)c\in(0,x)

Then I formed the inequality (as the sandwich lemma would probably be the easiest way to find such a limit),

0≤∣(−1)n+2(1+c)n+1(n+1)!xn+1∣≤∣(−1)n+2xn+1∣0 \le|\frac{(-1)^{n+2}}{(1+c)^{n+1}(n+1)!}x^{n+1}|\le|(-1)^{n+2}x^{n+1}|

But I don't think this has helped. So I ask, how should I go about solving this question? :hmmm:
(edited 11 years ago)

Scroll to see replies

Check your formula for R_{n,0}. Specifically, the derivative is wrong.
(edited 11 years ago)
Original post by IrrationalNumber
Check your formula for R_{n,0}. Specifically, the derivative is wrong.


But Rn,af(x)=f(n+1)(c)(n+1)!(x−a)n+1R_{n,a}f(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

So Rn,0f(x)=f(n+1)(c)(n+1)!xn+1R_{n,0}f(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}

f(x)=ln(1+x)f(x)=ln(1+x), x>−1x>-1.

f(n)(x)=(−1)n+1(1+x)nf^{(n)}(x)=\frac{(-1)^{n+1}}{(1+x)^{n}}

So f(n+1)(x)=(−1)n+2(1+x)n+1f^{(n+1)}(x)=\frac{(-1)^{n+2}}{(1+x)^{n+1}}

Surely? :hmmmm:
Original post by RamocitoMorales

f(n)(x)=(−1)n+1(1+x)nf^{(n)}(x)=\frac{(-1)^{n+1}}{(1+x)^{n}}

So f(n+1)(x)=(−1)n+2(1+x)n+1f^{(n+1)}(x)=\frac{(-1)^{n+2}}{(1+x)^{n+1}}

Surely? :hmmmm:

Check both of these.
Original post by IrrationalNumber
Check both of these.


Well f′(x)=11+xf'(x)=\frac{1}{1+x}

f′′(x)=−1(1+x)2f''(x)=-\frac{1}{(1+x)^2}

f′′′(x)=1(1+x)3f'''(x)=\frac{1}{(1+x)^3}

So f(n)(x)=(−1)n+1(1+x)nf^{(n)}(x)=\frac{(-1)^{n+1}}{(1+x)^{n}} as it holds true for all f(n)(x)f^{(n)}(x)

:lolwut:
Original post by RamocitoMorales
:

You'll kick yourself. What is the derivative of xn x^n ? So what about (1+x)n (1+x)^n for n∈Z n \in \mathbb{Z} ?
Original post by IrrationalNumber
You'll kick yourself. What is the derivative of xn x^n ? So what about (1+x)n (1+x)^n for n∈Z n \in \mathbb{Z} ?


Oh. :colondollar:

ddx(1+x)n=n(1+x)n−1\frac{d}{dx}(1+x)^n=n(1+x)^{n-1}.

Thanks.
(edited 11 years ago)
So the general solution would be,

f(n)(x)=(n−1)n+1(1+x)nf^{(n)}(x)=\frac{(n-1)^{n+1}}{(1+x)^{n}} for n>1n>1

Therefore Rn,0f(x)=nn+2(1+c)n+1(n+1)!xn+1R_{n,0}f(x)=\frac{n^{n+2}}{(1+c)^{n+1}(n+1)!}x^{n+1}
(edited 11 years ago)
Original post by RamocitoMorales
So the general solution would be

Do the first 2 or 3 explicitly, and you'll see why it's not that.
Original post by ghostwalker
Do the first 2 or 3 explicitly, and you'll see why it's not that.


Ah, right, what about,

f(n)(x)=(1−n)n+1(1+x)nf^{(n)}(x)=\frac{(1-n)^{n+1}}{(1+x)^{n}} for n>1n>1

I think that's right. But the fact it has to be for n>1n>1 is quite a nuisance, especially when the Maclaurin series will start at n=0n=0. What do I do? :sad:
Original post by RamocitoMorales
Ah, right, what about,

f(n)(x)=(1−n)n+1(1+x)nf^{(n)}(x)=\frac{(1-n)^{n+1}}{(1+x)^{n}} for n>1n>1

I think that's right. But the fact it has to be for n>1n>1 is quite a nuisance, especially when the Maclaurin series will start at n=0n=0. What do I do? :sad:


Still not right. Do the first 3 derivatives, make a guess, and try to prove it by induction. If your induction fails, then your guess is probably wrong.
Original post by IrrationalNumber
Still not right. Do the first 3 derivatives, make a guess, and try to prove it by induction. If your induction fails, then your guess is probably wrong.



f(n)(x)=(1−n)n+1(1+x)nf^{(n)}(x)=\frac{(1-n)^{n+1}}{(1+x)^{n}} for n>1n>1

For n=2n=2, we have f′′(x)=(1−2)2+1(1+x)2=(−1)3(1+x)2=−1(1+x)2f''(x)=\frac{(1-2)^{2+1}}{(1+x)^{2}}=\frac{(-1)^{3}}{(1+x)^{2}}=\frac{-1}{(1+x)^2}

Which is correct.

For n=3n=3, we have f′′′(x)=(1−3)3+1(1+x)3=(−2)4(1+x)3=...f'''(x)=\frac{(1-3)^{3+1}}{(1+x)^{3}}=\frac{(-2)^{4}}{(1+x)^{3}}=...

Oh, I see. :facepalm2:

I have a new general solution!

f(n)(x)=(−1)n+1⋅(n−1)(1+x)nf^{(n)}(x)=\frac{(-1)^{n+1}\cdot(n-1)}{(1+x)^{n}} for n>1n>1

Please tell me this one's right, I'm dying out here. :colondollar:

But I still have that problem with n having to be greater than 1. Damn it.
(edited 11 years ago)
Try that in the case n=4. Is it right?
Original post by IrrationalNumber
Try that in the case n=4. Is it right?


f(n)(x)=(−1)n+1⋅(n−1)(1+x)nf^{(n)}(x)=\frac{(-1)^{n+1}\cdot(n-1)}{(1+x)^{n}}

For n=4n=4, we have f(4)(x)=(−1)4+1⋅(4−1)(1+x)4=(−1)5⋅3(1+x)4=−1⋅3(1+x)4=−3(1+x)4f^{(4)}(x)=\frac{(-1)^{4+1}\cdot(4-1)}{(1+x)^{4}}=\frac{(-1)^{5}\cdot 3}{(1+x)^{4}}=\frac{-1\cdot 3}{(1+x)^{4}}=\frac{-3}{(1+x)^{4}}

Which I believe is correct.
Original post by RamocitoMorales

Which I believe is correct.

Work out, by hand, without trying to use your formula, what f(3)(x) f^(3)(x) is. Then differentiate and compare.
Original post by IrrationalNumber
Work out, by hand, without trying to use your formula, what f(3)(x) f^(3)(x) is. Then differentiate and compare.


By hand, I got f(3)(x)=2(1+x)3f^{(3)}(x)=\frac{2}{(1+x)^{3}}

Using my formula, I get f(3)(x)=(−1)3+1⋅(3−1)(1+x)3=(−1)4⋅2(1+x)3=1⋅2(1+x)3=2(1+x)3f^{(3)}(x)=\frac{(-1)^{3+1}\cdot(3-1)}{(1+x)^{3}}=\frac{(-1)^{4}\cdot 2}{(1+x)^{3}}=\frac{1\cdot 2}{(1+x)^{3}}=\frac{2}{(1+x)^{3}}

Which is the same as the calculation I made by hand.
Original post by RamocitoMorales
By hand, I got f(3)(x)=2(1+x)3f^{(3)}(x)=\frac{2}{(1+x)^{3}}

Using my formula, I get f(3)(x)=(−1)3+1⋅(3−1)(1+x)3=(−1)4⋅2(1+x)3=1⋅2(1+x)3=2(1+x)3f^{(3)}(x)=\frac{(-1)^{3+1}\cdot(3-1)}{(1+x)^{3}}=\frac{(-1)^{4}\cdot 2}{(1+x)^{3}}=\frac{1\cdot 2}{(1+x)^{3}}=\frac{2}{(1+x)^{3}}

Which is the same as the calculation I made by hand.


Right, but now differentiate that and compare with what your formula gives for n=4.
Original post by IrrationalNumber
Right, but now differentiate that and compare with what your formula gives for n=4.


Oh ****. :cry:

But I think I've got it this time...don't lose hope! :rofl:

f(n)(x)=(−1)n+1(n−1)!(1+x)nf^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{(1+x)^{n}}

And it works for n=1n=1 because I've defined 0!=10!=1

Spoiler

Original post by RamocitoMorales
Oh ****. :cry:

But I think I've got it this time...don't lose hope! :rofl:

I haven't!


f(n)(x)=(−1)n+1(n−1)!(1+x)nf^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{(1+x)^{n}}

And it works for n=1n=1 because I've defined 0!=10!=1

Would you like to have a go at proving it?


Spoiler


I disagree. Part of the way we come up with formulae is through trying to think of what the answer might be. The difference between maths and other experimental subjects is that we prove beyond a doubt that our guesses hold. So that is what you need to do now - prove it, rigorously. Intuitively, your answer looks right - from differentiating (1+x)^n we gain a factor of n, so the n+1th term will have n times the previous terms ie factorial.
IrrationalNumber
The difference between maths and other experimental subjects is that we prove beyond a doubt that our guesses hold.


As a 'philosophical sceptic' that's probably my motivation for studying it.

Original post by IrrationalNumber
Would you like to have a go at proving it?


Sure. I'll try to prove it by induction.

We have f(n)(x)=(−1)n+1(n−1)!(1+x)nf^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{(1+x)^{n}}

I've already shown it works for n=1n=1.

Assume it holds true for some constant n=kn=k, then for n=k+1n=k+1, we have

f(k+1)(x)=(−1)k+1+1(k+1−1)!(1+x)k+1=(−1)k+2k!(1+x)k+1=(−1)k⋅(−1)2k!(1+x)k+1=(−1)kk!(1+x)k+1f^{(k+1)}(x)=\frac{(-1)^{k+1+1}(k+1-1)!}{(1+x)^{k+1}}=\frac{(-1)^{k+2}k!}{(1+x)^{k+1}}=\frac{(-1)^k\cdot(-1)^{2}k!}{(1+x)^{k+1}}=\frac{(-1)^{k}k!}{(1+x)^{k+1}}

...then...is that sufficient proof as it's consistent with our formula for n+1n+1.

:dontknow:
(edited 11 years ago)

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