Exponential function proof problem

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SUMMARY

The forum discussion revolves around proving the exponential function using a limit approach. The proof involves the condition that for every natural number k, there exists a sequence of natural numbers n_k such that the sum of the sequence equals x, with the terms in non-increasing order. The critical conclusion is that if the limit of x_{k1} approaches zero, then the limit of the product (1+x_{k1})·...·(1+x_{k_{nk}}) converges to exp(x) or e^x. A key point raised is the necessity to establish that lim_{k→∞} n_{k}=∞ to validate the proof.

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rustyjoker
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Homework Statement



Big problem with exponential function proof assignment, need some help.

let
x≥0 and for every k[itex]\in N[/itex] there is [itex]n_{k}[/itex][itex]\in N[/itex] and

[itex]x_{k1}[/itex]≥...≥[itex]x_{k_{nk}}[/itex] and [itex]x_{k1}[/itex]+...+[itex]x_{k_{nk}}[/itex]=x.
Proof: if [itex]lim_{k→}∞ x_{k1}[/itex]=0 then [itex]lim_{k→}∞<br /> [/itex] (1+[itex]x_{k1}[/itex])·...·(1+[itex]x_{k_{nk}}[/itex])=exp(x)=[itex]e^{x}[/itex]
 
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rustyjoker said:

Homework Statement



Big problem with exponential function proof assignment, need some help.

let
x≥0 and for every k[itex]\in N[/itex] there is [itex]n_{k}[/itex][itex]\in N[/itex] and

[itex]x_{k1}[/itex]≥...≥[itex]x_{k_{nk}}[/itex] and [itex]x_{k1}[/itex]+...+[itex]x_{k_{nk}}[/itex]=x.
Proof: if [itex]lim_{k→}∞ x_{k1}[/itex]=0 then [itex]lim_{k→}∞<br /> [/itex] (1+[itex]x_{k1}[/itex])·...·(1+[itex]x_{k_{nk}}[/itex])=exp(x)=[itex]e^{x}[/itex]

There must be something wrong with the statement of hypotheses, because it allows me to take [itex]x_1 = x,\: x_2 = x_3 = \cdots = x_n = 0,[/itex] giving [itex]\lim_{n \rightarrow \infty} (1+x_1) \cdot (1+x_2) \cdots (1+x_n) = 1+x.[/itex]

RGV
 
Well, for an example if you think about the product
(1+0,009)(1+0,008)⋅...⋅(1+0,001)=1,045879514 and
exp(0,009+...+0,001)=1,046...
I think the idea is to proof that first [itex]lim_{k→}∞[/itex] [itex]n_{k}=∞, then <br /> [/itex] [itex]lim_{k→}∞[/itex] [itex]x_{k1}[/itex]=...=[itex]lim_{k→}∞[/itex] [itex]x_{k_{nk}}[/itex]=x/[itex]n_{k}[/itex]= 0. So we'd have
[itex]lim_{k→∞}[/itex] (1+x/[itex]n_{k}[/itex])[itex]^{n_{k}}[/itex] = [itex]e^{x}[/itex]
 

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