Question

What is f^7 (2) in this series?

Answer 1

f(x) = \[\sum_{n=0}^{\infty} (-1)^{n} \sqrt{n} (x-2)^{n} / n!\]

Answer 2

what is x - 2 if x = 2?

Answer 3

0... Yes I put in 0 but it didn't work...

Answer 4

Is f^7(x) the the seventh derivative (i.e. \(f^{(7)}(x)\)) of this series or \((f(x))^7\)?

Answer 5

yes

Answer 6

it is the seventh derivative

Answer 7

It is not zero because the 8th term of f(x) will be a constant when differentiated 7th time.

Answer 8

How do I know that? Mathematica!

Answer 9

Wait could you show me that?

Answer 10

is this the series

Answer 11

yes

Answer 12

$$
\Large
f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} \sqrt{n}~ (x-2)^{n}}{ n!}

$$

Answer 13

yup!!

Answer 14

$$
\Large {
f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} ~\sqrt{n}~ (x-2)^{n}}{ n!}
\\~\\
f ' (x) = \sum_{n=1}^{\infty} \frac{(-1)^{n} \sqrt{n}~\cdot n\cdot (x-2)^{n-1}}{ n!}
}
$$

Answer 15

lets use rational exponents

Answer 16

do you see the pattern emerging>?

Answer 17

\[
\begin{align*}
&\quad\sum_{n=0}^9 \frac{(-1)^{n}\sqrt{n}(x-2)^{n}}{ n!}\\
&=-(x-2)+\frac{(x-2)^2}{\sqrt{2}}-\frac{(x-2)^3}{2 \sqrt{3}}+\frac{1}{12}(x-2)^4\\
&\quad -\frac{(x-2)^5}{24 \sqrt{5}}+\frac{(x-2)^6}{120 \sqrt{6}}-\frac{(x-2)^7}{720
\sqrt{7}}+\frac{(x-2)^8}{10080 \sqrt{2}}-\frac{(x-2)^9}{120960}
\end{align*}
\]
After differentiating it 7 times, the first 7 terms vanishes, the 8th term will become a constant and the 9+ term will have (x-2) as a term. Hence, you only have to differentiate the 8th term 7 times to get \(f^{(7)}(2)\).

Answer 18

Wait no! The 7th term will become a constant and the 8+ terms will have (x-2) as a term.

Answer 19

do you see why it canceled?

Answer 20

$$
\Large {
f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} ~{n}^{1/ 2}~ (x-2)^{n}}{ n!}
\\~\\\text{power rule}\\
f ' (x) = \sum_{n=1}^{\infty} \frac{(-1)^{n} ~{n}^{1/ 2}~\cdot n\cdot (x-2)^{n-1}}{ n!}
\\~\\=\sum_{n=1}^{\infty} \frac{(-1)^{n} ~{n}^{1/ 2}~\cdot n\cdot (x-2)^{n-1}}{ n\cdot (n-1)!}
\\~\\ =\sum_{n=1}^{\infty} \frac{(-1)^{n} ~{n}^{1/ 2}~\cdot \cancel n\cdot (x-2)^{n-1}}{ \cancel n\cdot (n-1)!}

\\=\sum_{n=1}^{\infty} \frac{(-1)^{n} ~{n}^{1/ 2} (x-2)^{n-1}}{ (n-1)!}

}$$

Answer 21

because
n! = n (n-1)(n-2)...(3)(2)(1) = n (n-1)!

Answer 22

Ok I understand that part...

Answer 23

But I don't understand thomas' explanation...

Answer 24

thomas expanded the series itself
then take the derivative 7 times, terms will vanish

Answer 25

Oh I see

Answer 26

I understand now, thank you :)

Answer 27

Or put it this way.
\[
\begin{align*}
f(x)&=\sum_{n=0}^\infty \frac{(-1)^{n} \sqrt{n}}{ n!}(x-2)^{n}\\
&=\sum_{n=0}^\infty c_n(x-2)^{n}\qquad c_n=\frac{(-1)^n\sqrt{n}}{n!}\\
&=c_1(x-2)+c_2(x-2)^2+c_3(x-3)^3+\cdots\qquad c_0=0\\
f^{(7)}(x)&=7!c_7+(x-2)sth\\
f^{(7)}(2)&=\frac{7!(-1)^7\sqrt{7}}{7!}=-\sqrt{7}
\end{align*}
\]

Answer 28

one thing we might want to do is change the lower limit, because its a bit odd to start at n = 7

Answer 29

if we do it the other way we get an indeterminate expression 0^0
$$
\Large {


f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} ~{n}^{1/ 2}~ (x-2)^{n}}{ n!}
\\~\\f^{(7)}(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1} ~{(n+7)}^{1/ 2}(x-2)^{n}}{n!}
\\~\\f^{(7)}(2) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1} ~{(n+7)}^{1/ 2}(x-2)^{n}}{n!}
}$$

Answer 30

i'm a bit confused as to which expression it should be, can you take a screenshot of the original problem

Answer 31

i guess we can allow that 0^0 = 1 , in this series.
for instance we allow it in other cases
$$ \Large {e^x = \sum_{n=0}^{\infty } \frac{x^n}{n!}
\\~\\ e^0 = \frac{0^0}{0!} + \frac{0^1}{1!} + \frac{0^2}{2!} + ...

}$$

Answer 32

ok then your work and my work makes sense

Answer 33

Can we justify \(0^0\)=1 by saying \(\lim_{x\to 0}x^0=1\)?

Answer 34

\[
\Large {


f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} ~{n}^{1/ 2}~ (x-2)^{n}}{ n!}
\\~\\f^{(7)}(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1} ~{(n+7)}^{1/ 2}(x-2)^{n}}{n!}
\\~\\f^{(7)}(2) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1} ~{(n+7)}^{1/ 2}(x-2)^{n}}{n!}
\\~\\ = \frac{(-1)^1\cdot 7^{1/2} \cdot 0^0}{0!} +0 + 0 + ...
}\]

Answer 35

yes that makes sense, we can plug in the discontinuity

Answer 36

but strictly speaking it is indeterminate , i believe

Answer 37

what if you did instead
lim 0^x , as x->0 , then you would have 0

Answer 38

$$
\Large{
\lim_{x\to 0} ~0^x = 0
\\~\\
\lim_{x\to 0}~ x^0 = 1
}
$$

Answer 39

I think \(0^0\) is not defined precisely because of this reason.

Answer 40

but in the series we see that 0^0 has to be equal to 1, as we see in the series e^0

Answer 41

\[ \large {e^x = \sum_{n=0}^{\infty } \frac{x^n}{n!}
\\~\\ e^0 = \frac{0^0}{0!} + \frac{0^1}{1!} + \frac{0^2}{2!} + ...
=\frac{0^0}{1 } + 0 + 0 + ... = \frac{0^0}{1 }
\\~\\e^0 = \frac{0^0}{1 }
\\~\\1= \frac{0^0}{1 }
\\~\\1\cdot 1 = 0^0
}\]

Answer 42

we might as well add another limit
\[
\Large{
\lim_{x\to 0} ~0^x = 0
\\~\\ \lim_{x\to 0}~ x^0 = 1
\\~\\ \lim_{x\to 0^{+}}~ x^x = 1

}\]

Answer 43

interestingly enough my software computes 0^0 as 1
http://prntscr.com/76bw11

but my calculator gives me error

Answer 44

Can we just say \(\lim_{(x,y)\to(0,0)}x^y\) does not exist?

Answer 45

right, that combines the two cases above

Answer 46

and when x = y, the limit is 1

Answer 47

wolfram
https://www.wolframalpha.com/input/?i=0%5E0

Answer 48

err, actually three cases you can use to show limit does not exist, but two different paths are sufficient,