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Suppose I have a function $f(x)=x$

And I can rewrite the function as follows :

$$x=e^{\ln{x}}$$

Rewrite it again in Maclaurin series form:

$$e^{\ln{x}}=\sum_{n=0}^{\infty} \frac{\ln^n{x}}{n!}$$

So we have

$$x=\sum_{n=0}^{\infty} \frac{\ln^n{x}}{n!}$$

When I plug $x=1$ into the equation, then:

$$\begin{align} 1&=\sum_{n=0}^{\infty} \frac{\ln^n{1}}{n!}\\ &=\sum_{n=0}^{\infty} \frac{(\ln{1})^n}{n!}\\ &=\sum_{n=0}^{\infty} \frac{0^n}{n!}\\ &=\sum_{n=0}^{\infty} \frac{0}{n!}\\ &=\sum_{n=0}^{\infty} 0\\ &=0\\ \\ \therefore 1&=0 \end{align}$$

Why this can be happen?

Please tell me. Thanks.

Martin Sleziak
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user516076
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2 Answers2

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Power series such as $\sum_{n=0}^\infty a_nT^n$ work under the assumption that $T^0$ stands for the constant $1$ function. Therefore the RHS is $1$, not $0$.

  • I still don't understand. Please spot the mistake on my proof. Thanks. – user516076 Nov 30 '19 at 10:11
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    The mistake is saying that $\sum_{n=0}^\infty \frac{(\ln1)^n}{n!}=0$ –  Nov 30 '19 at 10:16
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    @usrr516076 We have $0^0=1$. You thought $0^0=0$. That is the mistake that this answer points out. – Arthur Nov 30 '19 at 10:35
  • @Arthur So $0^0=1$? before this, i thought that $0^0$ is undefined considering the limit $x^x$ from $0^+$ and $0^-$ is not same, bcz of in $0^-$ doesn't exist? This is one of the new things that I just learned in math. Thanks anyway. Could we use this fact $0^0=1$ forever? – user516076 Nov 30 '19 at 10:58
  • @user516076 There are many instances where the preferred convention for exponents which are implicitly in $\Bbb N$ is $0^n=\begin{cases}1&\text{if }n=0\ 0&\text{if }n>0\end{cases}$. –  Nov 30 '19 at 11:04
  • @user516076 Limits are not the only way to justify a convention. "It is extremely convenient with basically no downsides" is another one, which applies to $0^0=1$. – Arthur Nov 30 '19 at 11:06
  • @Gae.S. if as you said $0^0=1$ then could it be $0^0=0^{1-1}=\frac{0}{0}=1$? (I'm just honestly asking). It confuse me. For this reason, somehow i assumed $0^0$ is undefined? Thanks for the reply anyway – user516076 Nov 30 '19 at 11:24
  • @user516076 There is an identity $a^{n-m}=\frac{a^n}{a^m}$ which works mainly for all $a\ne 0$ and $n,m\in\Bbb Z$, and for all $a\in(0,\infty)$ and $n,m\in\Bbb R$. It may not extend to other values. Notice that your alleged issue wth $0^{1-1}$ is the same that you have for $0^{2-1}$. –  Nov 30 '19 at 11:32
  • @user516076 $0^0$ by itself is usually undefined since $0^x=0$ but $x^0=1$ so you'd have a contradiction. But for convenience's sake (and because we're interested in the case of $x^0$, not $0^x$), it's simpler to define $0^0$ in the Taylor series to be $1$. This might seem like it causes a contradiction but if we're careful (e.g. by defining $0^0$ as a limit of $x^0$) it's solid. This is covered further in Question 11150. – Jam Nov 30 '19 at 11:56
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The Maclaurin development of the exponential is

$$\exp(z)=1+z+\frac{z^2}2+\frac{z^3}{3!}+\cdots=1+\sum_{n=1}^\infty\frac{z^n}{n!}$$ so that $\exp(0)=1$.

Technically, $\dfrac{z^0}{0!}$ is not well defined at $z=0$, safer to write $1$.

  • $e^0=1$ yes, i agree with that. But in summation form? – user516076 Nov 30 '19 at 10:15
  • Reminds me a lot of how, more generally, we might write a polynomial $p$ as

    $$p(x) = \sum_{k=0}^n a_k x^k$$

    but then $p(0)$ involves a $0^0$ term, technically. Of course, I imagine whenever that happens there's the implied statement "$0^0=1$ on evaluation of the constant term."

     – PrincessEev Nov 30 '19 at 10:16
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    @user516076: in fact it was never said that the first term of Taylor is $f(x)x^0$. It is $f(0)$. Your "summation form" is just undefined. Worse, you make it invalid by your interpretation of $x^0$. –  Nov 30 '19 at 10:16
  • @YvesDaoust So, in short i can't rewrite $1$ as that sum, involving the maclaurin series of $e^{\ln{x}}$?. Bcz, when i checked on wolfram alpha, it said the sum doesn't exist due to singularity at one or more evaluation points. But, it works for $x=2$ – user516076 Nov 30 '19 at 10:26
  • @user516076: it doesn't work at $x=0$, which is not your case, you plugged $x=1$, didn't you ? Regarding the sum , I think I explained enough. I changed my notation to reduce ambiguity. –  Nov 30 '19 at 10:34