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- Thread starter georg gill
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- #2

Stephen Tashi

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without using derivation rules for polynomials?

It isn't clear what you mean.

Are you asking for a formal proof instead the informal way of doing things? The informal way assumes that function has a power series and says it's plausible that the coefficients can be found by doing a term-by-term differentiations of the power series. Is "using derivation rules for polynomials" your terminology for the informal method?

- #3

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sorry without using the fact that

I:

[tex] \frac{d}{dx}x^n=nx^{n-1}[/tex]

I belive this explains taylor theorem

http://bildr.no/view/1030479

but it uses the rule I mentioned above

I:

[tex] \frac{d}{dx}x^n=nx^{n-1}[/tex]

I belive this explains taylor theorem

http://bildr.no/view/1030479

but it uses the rule I mentioned above

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- #4

Stephen Tashi

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Stephen Tashi

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I belive this explains taylor theorem

http://bildr.no/view/1030479

but it uses the rule I mentioned above

I'll agree with calling that an explanation, but it isn't a formal proof.

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This one is the closest proof I have found but it uses the fact also

http://en.wikipedia.org/wiki/Taylor's_theorem#Proof_for_Taylor.27s_theorem_in_one_real_variable

wonder if it possible to prove it with L'hopitals without using the fact I mentioned above

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- #7

Stephen Tashi

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I guess the same problem goes for binomial theorem?

You should polish your writing skills, georg! Are you asking whether there is a proof of Taylor's formula that does not rely on binomial theorem? Or are you asking whether there is a proof of the binomial theorm that does not rely on the differentiation rules for polynomials? I think the binomial theorem can be proven without using derivatives.

- #8

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You should polish your writing skills, georg! Are you asking whether there is a proof of Taylor's formula that does not rely on binomial theorem? Or are you asking whether there is a proof of the binomial theorm that does not rely on the differentiation rules for polynomials? I think the binomial theorem can be proven without using derivatives.

Sorry I will try to write more correct

I meant if there were a proof for binomal theorem that did not use

I:

[tex] \frac{d}{dx}x^n=nx^{n-1}[/tex]

if so it would be great to have a proof for all real numbers for the binomial theorem that does not use the fact I

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- #9

Stephen Tashi

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It isn't exactly clear to me what you mean by a proof. As I said, the link you gave had an informal explanation of Taylor's series, but it was hardly a proof. So I can't figure out exactly what your are asking. Can you explain why your are interested in an explanation that avoids the differentiation rule? Are you trying to explain the expansion of [itex] (x + a)^r[/itex] to some students who have not yet studied calculus?

- #10

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I will try to explain it clear

The reason is in these links wich are my own notes:

http://bildr.no/view/1031000

In link above the problem is to prove that

[tex](a^y)^{\frac{1}{C}}=a^{\frac{y}{C}}[/tex]

which is not understandable for y=1.23 and C=3.12 for example

Then I tried to proove 4 in the first link by proving logrule a different way shown here in a second link:

http://bildr.no/view/1031585

But as it says in the second link I get stuck with proof for differentiating polynomials so I tried to prove how to differentiate polynomials another way and I found out it could be proved by binomial theorem. So I wanted to prove binomial theorem for all real numbers without using rule for differentiating a polynomial (I also thought binomial theorem could be proved by maclaurin series that is why I asked about proof for Taylor series. It seemed to make sense for me since binomial theorem in my book is derieved from taylor series with a=0)

I have asked about this before but I thought taylor polynomial wold be a good possibility and I had almost ruled out binomial theorem but if I can use binomial theorem to prove how to differentiate polynomials for all real numbers it would be great

The reason is in these links wich are my own notes:

http://bildr.no/view/1031000

In link above the problem is to prove that

[tex](a^y)^{\frac{1}{C}}=a^{\frac{y}{C}}[/tex]

which is not understandable for y=1.23 and C=3.12 for example

Then I tried to proove 4 in the first link by proving logrule a different way shown here in a second link:

http://bildr.no/view/1031585

But as it says in the second link I get stuck with proof for differentiating polynomials so I tried to prove how to differentiate polynomials another way and I found out it could be proved by binomial theorem. So I wanted to prove binomial theorem for all real numbers without using rule for differentiating a polynomial (I also thought binomial theorem could be proved by maclaurin series that is why I asked about proof for Taylor series. It seemed to make sense for me since binomial theorem in my book is derieved from taylor series with a=0)

I have asked about this before but I thought taylor polynomial wold be a good possibility and I had almost ruled out binomial theorem but if I can use binomial theorem to prove how to differentiate polynomials for all real numbers it would be great

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- #11

Stephen Tashi

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Ok I will try to explain it as clear as possible:)

Purpose:

Prove that

[tex](e^x)^y=e^{xy}[/tex] (a)

See my first link for a way to show (a)

First link:

http://bildr.no/view/1031000 I

As I said:

the problem in the first link is to prove that

[tex](a^y)^{\frac{1}{C}}=a^{\frac{y}{C}}[/tex]

which is not understandable for y=1.23 and C=3.12 for example how could one say that

[tex](a^{1.23})^{\frac{1}{3.12}}=a^{\frac{1.23}{3.12}}=a^{\frac{41}{104}}[/tex]

So I tried another way:

Here in link two:

http://bildr.no/view/1031585 II

Proof of chain rule is proved by linearization shown in this link uploaded on scribd.com:

http://www.scribd.com/doc/73314666/Proof-Chain-Rule-English-Ver-PDF [Broken]

Then I have to prove

[tex]\frac{d}{dx}e^x=e^x[/tex]

First I would find the derivative of lnx because of inverse relationship (this way is how it is described in my book)

http://bildr.no/view/946470

Then I use that to find the derivative of [tex]\frac{d}{dx}e^x=e^x[/tex]:

http://bildr.no/view/1026637

So it is the problem with how to differentiate polynomials in link two (II) that I can't prove here

So I tried to prove how to differentiate polynomials another way and I found out it could be proved by binomial theorem. So I wanted to prove binomial theorem for all real numbers without using rule for differentiating a polynomial. Is that possible?

Is this clear formulation? If it is not I don't know where I am not being clear :(

Purpose:

Prove that

[tex](e^x)^y=e^{xy}[/tex] (a)

See my first link for a way to show (a)

First link:

http://bildr.no/view/1031000 I

As I said:

the problem in the first link is to prove that

[tex](a^y)^{\frac{1}{C}}=a^{\frac{y}{C}}[/tex]

which is not understandable for y=1.23 and C=3.12 for example how could one say that

[tex](a^{1.23})^{\frac{1}{3.12}}=a^{\frac{1.23}{3.12}}=a^{\frac{41}{104}}[/tex]

So I tried another way:

Here in link two:

http://bildr.no/view/1031585 II

Proof of chain rule is proved by linearization shown in this link uploaded on scribd.com:

http://www.scribd.com/doc/73314666/Proof-Chain-Rule-English-Ver-PDF [Broken]

Then I have to prove

[tex]\frac{d}{dx}e^x=e^x[/tex]

First I would find the derivative of lnx because of inverse relationship (this way is how it is described in my book)

http://bildr.no/view/946470

Then I use that to find the derivative of [tex]\frac{d}{dx}e^x=e^x[/tex]:

http://bildr.no/view/1026637

So it is the problem with how to differentiate polynomials in link two (II) that I can't prove here

So I tried to prove how to differentiate polynomials another way and I found out it could be proved by binomial theorem. So I wanted to prove binomial theorem for all real numbers without using rule for differentiating a polynomial. Is that possible?

Is this clear formulation? If it is not I don't know where I am not being clear :(

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- #13

Stephen Tashi

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I don't know where I am not being clear :(

You have described a theorem that you wish to prove. However, you have not completely described the body of mathematics that you have assumed or proven before reaching this theorem. For example, if you are studying from a textbook, what theorems have been proven? What definitions have been given? There are many different ways to develop mathematics. Different sets of assumptions can be used. Theorems can be proven in different chonological orders.

You still have not stated your purpose in doing this.

- #14

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I am sorry this is above me is it possible to be specific in where it is not right?

I also tried to specify the purpose

why is it not specified or what is wrong? For me it would be nice to prove this or is puropse something else?

I also tried to specify the purpose

Ok I will try to explain it as clear as possible:)

Purpose:

Prove that

[tex](e^x)^y=e^{xy}[/tex]

why is it not specified or what is wrong? For me it would be nice to prove this or is puropse something else?

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- #15

Stephen Tashi

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The first is why you titled the thread "proof taylor formula" and why are you mentioning the "binomal theorem"?

As I understand your goal, you wish to prove:

Theorem 1: If [itex] x [/itex] is a real number and [itex] r [/itex] is a rational number then [itex](e^x)^r = e^{xr} [/itex].

Is that correct? Or do you want [itex] x [/itex] to also be a rational number?

This theorem is not called "the binomial theorem" and it is not "taylors formula".

When the properties of the real number system are developed in advanced textbooks the theorem that is proven is:

Theorem 2: If [itex] x , r[/itex] and [itex] A [/itex] are each real numbers and [itex] A \ge 0 [/itex] then [itex] (A^x)^r = A^{xr} [/itex].

In advanced texts, Theorem 2 is proven without resorting to derivatives or logarithms. It may require using theorems about limits. It doesn't require defining the special number [itex] e [/itex].

It is very arduous to prove Theorem 2. As I recall, one first establishes properties of the integers, then the rational numbers are defined in terms of the integers. Then the irrational numbers are defined as "Dedkind Cuts" of the rational numbers.

You want is to prove Theorem 1 without assuming Theorem 2. if you don't assume Theorem 2, what properties of the real number system are you willing to assume?

You are apparently using a calculus textbook, not an advanced textbook that develops the properties of the real number system in a rigorous manner. The elementary properties of the real number are a prerequisite for the definitions and theorems used in algebra and calculus. If you use calculus or algebra to "prove" an elementary property of the real numbers, you are guilty of using circular reasoning.

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As I understand your goal, you wish to prove:

Theorem 1: If [itex] x [/itex] is a real number and [itex] r [/itex] is a rational number then [itex](e^x)^r = e^{xr} [/itex].

Is that correct? Or do you want [itex] x [/itex] to also be a rational number?

I would like that both x and r would be real numbers.

Could you tell me the name of a book which has the proof for theorem 2 that you mention? I would in any case really like to have a proof of it.

And as for my attempt to get to a proof for

[itex](e^x)^r = e^{xr} [/itex].

I thought whether you used e as base or a unknown base A the rules are the same so for simplicity I copied and pasted proof from book also to make sure I did not misspell anything from the book of importance. So I guess my question is if my explanation is understandable if one takes into account that logrules works for any base.

Yes I guess thread is messy. Sorry for that. I asked about taylor theorem first because binomial theorem is proved by Taylor theorem

But now I have to look more on taylor series to be sure if I could explain that one could use taylor theorem directly to prove how to differentiate a polynomial. Thank you for the help. Hopefully I can ask about that later

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- #17

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Ok this was my thought for using taylorpolynomial as approximation for

[tex]f(x)=(h+x)^2[/tex]

around x=0 (guess it is a good place to approximate around since slope of polynomials often are the same in both directions from x=0)

maclaurinseries

[tex]f(0)+f^{\prime} (0)(x)+\frac {f^{(2)}(0)}{2 !} (x)^2+...{f^{(n)}(0)}{n !} (x)^n[/tex]

becomes

[tex]h^n+n(h)^{n-1}x+\frac{n(n-1)(h)^{n-2}}{2 !} (x)^2+...\frac{n(n-1)....(n-(n+1))}{n !} (x)^n[/tex]

which when n is large enough becomes a good approximation for f(x)

In this link it seems they have used maclaurin series to express

[tex]f(x)=(h+x)^2[/tex]

in the first rule rule for how to derivate polynomials:

http://www.wyzant.com/Help/Math/Calculus/Derivative_Proofs/Power_Rule.aspx

That was why I wanted to prove taylor polynomials without using the rule for how to differentiate a polynomial and use it in my struggles above:

[tex]f(x)=(h+x)^2[/tex]

around x=0 (guess it is a good place to approximate around since slope of polynomials often are the same in both directions from x=0)

maclaurinseries

[tex]f(0)+f^{\prime} (0)(x)+\frac {f^{(2)}(0)}{2 !} (x)^2+...{f^{(n)}(0)}{n !} (x)^n[/tex]

becomes

[tex]h^n+n(h)^{n-1}x+\frac{n(n-1)(h)^{n-2}}{2 !} (x)^2+...\frac{n(n-1)....(n-(n+1))}{n !} (x)^n[/tex]

which when n is large enough becomes a good approximation for f(x)

In this link it seems they have used maclaurin series to express

[tex]f(x)=(h+x)^2[/tex]

in the first rule rule for how to derivate polynomials:

http://www.wyzant.com/Help/Math/Calculus/Derivative_Proofs/Power_Rule.aspx

That was why I wanted to prove taylor polynomials without using the rule for how to differentiate a polynomial and use it in my struggles above:

So it is the problem with how to differentiate polynomials in link two (II) that I can't prove here

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- #18

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[tex]f(x+h)={A}_{0}+{A}_{1}h+{A}_{2}{h}^{2}+{A}_{3}{h}^{3}+...[/tex]

Take the limit as h approaches 0.

[tex]f(x)={A}_{0}[/tex]

[tex]f(x+h)=f(x)+{A}_{1}h+{A}_{2}{h}^{2}+{A}_{3}{h}^{3}+...[/tex].

Take the derivative with respect to h.

[tex]f'(x+h)={A}_{1}+2{A}_{2}h+3{A}_{3}{h}^{2}+4{A}_{4}{h}^{3}+...[/tex]

Take the limit as h approaches 0.

[tex]f'(x)={A}_{1}[/tex]

[tex]f'(x+h)=f'(x)+2{A}_{2}h+3{A}_{3}{h}^{2}+4{A}_{4}{h}^{3}+...[/tex]

Take the derivative with respect to h.

[tex]f''(x+h)=2{A}_{2}+6{A}_{3}h+12{A}_{4}{h}^{2}+...[/tex]

Take the limit as h approaches 0.

[tex]f''(x)=2{A}_{2}[/tex]

You can repeat this a few times to get a good picture of what is going on.

But you will find that

[tex]{f}^{(n)}(x)=n!{A}_{n}[/tex]

Substitute into our original power series.

[tex]f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...[/tex]

[tex]f(x+h)=\sum_{k=0}^{\infty}\frac{{h}^{n}}{n!}{f}^{(n)}(x)[/tex]

[tex]f(x+h)=\sum_{k=0}^{\infty}\frac{{h}^{n}}{n!}\frac{{\mbox{d}}^{n}f}{{\mbox{d}x}^{n}}[/tex]

But I had to use the elementary power rule. But why do you not want to use it? You can easily prove it.

[tex]y=x^n[/tex]

[tex]\mbox{ln}y=n\mbox{ln}x[/tex]

[tex]\frac{1}{y}\frac{\mbox{d}y}{\mbox{d}x}=\frac{n}{x}[/tex]

[tex]\frac{\mbox{d}y}{\mbox{d}x}=\frac{nx^n}{n}[/tex]

[tex]\frac{\mbox{d}y}{\mbox{d}x}=n{x}^{n-1}[/tex]

Take the limit as h approaches 0.

[tex]f(x)={A}_{0}[/tex]

[tex]f(x+h)=f(x)+{A}_{1}h+{A}_{2}{h}^{2}+{A}_{3}{h}^{3}+...[/tex].

Take the derivative with respect to h.

[tex]f'(x+h)={A}_{1}+2{A}_{2}h+3{A}_{3}{h}^{2}+4{A}_{4}{h}^{3}+...[/tex]

Take the limit as h approaches 0.

[tex]f'(x)={A}_{1}[/tex]

[tex]f'(x+h)=f'(x)+2{A}_{2}h+3{A}_{3}{h}^{2}+4{A}_{4}{h}^{3}+...[/tex]

Take the derivative with respect to h.

[tex]f''(x+h)=2{A}_{2}+6{A}_{3}h+12{A}_{4}{h}^{2}+...[/tex]

Take the limit as h approaches 0.

[tex]f''(x)=2{A}_{2}[/tex]

You can repeat this a few times to get a good picture of what is going on.

But you will find that

[tex]{f}^{(n)}(x)=n!{A}_{n}[/tex]

Substitute into our original power series.

[tex]f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...[/tex]

[tex]f(x+h)=\sum_{k=0}^{\infty}\frac{{h}^{n}}{n!}{f}^{(n)}(x)[/tex]

[tex]f(x+h)=\sum_{k=0}^{\infty}\frac{{h}^{n}}{n!}\frac{{\mbox{d}}^{n}f}{{\mbox{d}x}^{n}}[/tex]

But I had to use the elementary power rule. But why do you not want to use it? You can easily prove it.

[tex]y=x^n[/tex]

[tex]\mbox{ln}y=n\mbox{ln}x[/tex]

[tex]\frac{1}{y}\frac{\mbox{d}y}{\mbox{d}x}=\frac{n}{x}[/tex]

[tex]\frac{\mbox{d}y}{\mbox{d}x}=\frac{nx^n}{n}[/tex]

[tex]\frac{\mbox{d}y}{\mbox{d}x}=n{x}^{n-1}[/tex]

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- #19

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But I had to use the elementary power rule. But why do you not want to use it? You can easily prove it.

[tex]y=x^n[/tex]

[tex]\mbox{ln}y=n\mbox{ln}x[/tex]

[tex]\frac{1}{y}\frac{\mbox{d}y}{\mbox{d}x}=\frac{n}{x}[/tex]

[tex]\frac{\mbox{d}y}{\mbox{d}x}=\frac{nx^n}{n}[/tex]

[tex]\frac{\mbox{d}y}{\mbox{d}x}=n{x}^{n-1}[/tex]

You use the rule for log with any base:

[tex]log_x a^x=x log_x a[/tex]

I wanted to prove 4 in link here

http://bildr.no/view/1031000 (t)

there i used the log rule you used so I had to prove that one as well. I tried proving it with derivation:

http://bildr.no/view/1031585

but it relies on among others rules rule for differntiation of polynomials. The other rules I can prove but the proof for power rule relies on log rule and then I can't prove the log rule this way.

I have tried to explain it more clear in post number 12 on the first page of this thread

Someone said earlier in this thread that it was a proof for 4 (4 is in the first link (t) in this post) that used among other things dedekinds cut to prove it for all real numbers if someone know where I could find it or buy it online I would be very thankful!

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- #20

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You use the rule for log with any base:

[tex]log_x a^x=x log_x a[/tex]

I wanted to prove 4 in link here

http://bildr.no/view/1031000 (t)

there i used the log rule you used so I had to prove that one as well. I tried proving it with derivation:

http://bildr.no/view/1031585

but it relies on among others rules rule for differntiation of polynomials. The other rules I can prove but the proof for power rule relies on log rule and then I can't prove the log rule this way.

I have tried to explain it more clear in post number 12 on the first page of this thread

Someone said earlier in this thread that it was a proof for 4 (4 is in the first link (t) in this post) that used among other things dedekinds cut to prove it for all real numbers if someone know where I could find it or buy it online I would be very thankful!

[tex]y=\mbox{ln}x[/tex]

[tex]x=\exp y [/tex]

[tex]\frac{\mbox{d}x}{\mbox{d}y}=\exp y [/tex]

[tex]\frac{\mbox{d}y}{\mbox{d}x}=\frac{1}{\exp y} [/tex]

[tex]\frac{\mbox{d}y}{\mbox{d}x}=\frac{1}{\exp\mbox{ln}x}[/tex]

[tex]\frac{\mbox{d}y}{\mbox{d}x}=\frac{1}{x}[/tex]

And I have used [tex]\frac{\mbox{d}}{\mbox{d}x} \exp x =\exp x [/tex]. The proof of that is

[tex]\frac{\mbox{d}}{\mbox{d}x}\exp x =\lim_{h\rightarrow 0}\frac{\exp(x+h)-\exp(x)}{h}[/tex]

[tex]\frac{\mbox{d}}{\mbox{d}x}\exp x =\exp(x)\lim_{h\rightarrow 0}\frac{\exp(h)-1}{h}[/tex]

[tex]\mbox{We know that } e=\lim_{n\rightarrow\infty}{\left(1+\frac{1}{n} \right)}

^{n}=\lim_{h \rightarrow 0}\left(1+h\right)^\frac{1}{h}[/tex]

[tex]\frac{\mbox{d}}{\mbox{d}x}\exp x =\exp(x)\lim_{h\rightarrow 0}\frac{{\sqrt[h]{1+h}}^{h}-1}{h}[/tex]

[tex]\frac{\mbox{d}}{\mbox{d}x}\exp x =\exp(x)\lim_{h\rightarrow 0}\frac{1+h-1}{h}[/tex]

[tex]\frac{\mbox{d}}{\mbox{d}x}\exp x =\exp(x)[/tex]

- #21

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Very cool proves for derivation of lnx and

[tex]e^x[/tex]

So I guess that my proof should work if only I could proove derivation of polynomial in a way that fits the proof then again who knows maybe a different proof for

[tex](e^x)^y=e^{xy}[/tex]

is the way to go

and maybe it is so that the way that I explained 2 posts above will not be provable

[tex]e^x[/tex]

So I guess that my proof should work if only I could proove derivation of polynomial in a way that fits the proof then again who knows maybe a different proof for

[tex](e^x)^y=e^{xy}[/tex]

is the way to go

and maybe it is so that the way that I explained 2 posts above will not be provable

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- #22

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[tex]\mbox{We know that } e=\lim_{n\rightarrow\infty}{\left(1+\frac{1}{n} \right)}

^{n}=\lim_{h \rightarrow 0}\left(1+h\right)^\frac{1}{h}[/tex]

And I have proof for this one in my book :)

http://bildr.no/view/1034162

I need to prove lhopitals I am working on it

- #23

HallsofIvy

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One important example is [itex]f(x)= e^{-1/x^2}[/itex] if [itex]x\ne 0[/itex], f(0)= 0. It can be shown that f is infinitely differentiable, and that its derivatives of any order, at x= 0, are 0. That means that its MacLauren series (Taylor series at x= 0) is just the constant 0. But f is clearly not 0 any where except at x= 0.

- #24

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my attempt to prove L'hopital's. Please feel free to say something if something is unclear:)

http://www.keepandshare.com/doc/3298776/bevis-l-hopital-s-0-over-0-english-pdf-pdf-november-24-2011-2-32-pm-481k?da=y [Broken]

In this proof for how to differentiate

[tex]e^x[/tex]

I think I have found an easier approach to proove L'hopital's without using unnecessary facts shown in the bottom here. This is an almost full proof of proving how to differentiate [tex]e^x[/tex]

http://www.viewdocsonline.com/document/wldxrc

http://www.keepandshare.com/doc/3298776/bevis-l-hopital-s-0-over-0-english-pdf-pdf-november-24-2011-2-32-pm-481k?da=y [Broken]

In this proof for how to differentiate

[tex]e^x[/tex]

I think I have found an easier approach to proove L'hopital's without using unnecessary facts shown in the bottom here. This is an almost full proof of proving how to differentiate [tex]e^x[/tex]

http://www.viewdocsonline.com/document/wldxrc

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- #25

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I need to prove lhopitals I am working on it

Well, L'Hospital's rule is obvious if you look at it immediately after looking at Cauchy's mean value theorem.

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