Prove (or disprove): Let $f_n[g(x)]$ denote the $n^\text{th}$ derivative of $g(x)$ at $x=1$ , then $f_n[^n(x)]=f_n[^{n+1}(x)]=f_n[^{n+2}(x)]=\ldots$ for all positive integers $n$ .

https://brilliant.org/problems/separation-at-tinfty/

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#Calculus

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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Great observation! It's easy to see why this would be true. Let $T_n(f(x))$ be the Taylor polynomial of $f(x)$ at $x=0$ .

If $T_n(f(x))=T_n(g(x))$ for two functions, then $T_{n+1}(\ln(x+1)f(x))=T_{n+1}(\ln(x+1)g(x))$ ; since $\ln(x+1)$ does not have a constant term, the degree gets "pushed up". This in turn implies that $T_{n+1}((x+1)^{f(x)})=T_{n+1}((x+1)^{g(x)})$ , by exponentiation.

Now $T_1(x+1)=T_1\left((x+1)^{x+1}\right)$ , by inspection ; applying the above observation $n-1$ times, we find that $T_n\left(^n(x+1)\right)=T_{n}\left(^{n+1}(x+1)\right)$ and therefore $T_n\left(^n(x+1)\right)=T_{n}\left(^m(x+1)\right)$ for all $m\geq{n}$ .

Let $D_n(f(x))$ be the nth derivative of $f(x)$ at $x=0$ . Since $D_n$ is determined by $T_n$ , we have $D_n\left(^n(x+1)\right)=D_{n}\left(^m(x+1)\right)$ for all $m\geq{n}$ .

In particular, $D_3\left(^7(x+1)\right)=D_{3}\left(^3(x+1)\right)=9$ , as we saw in Pi Han Goh's earlier problem.

Otto Bretscher - 6 years, 1 month ago

HEY HEY HEY, don't display that last line, you're giving out the answers for free!

You take the words out of my mouth. Haha, just kidding. Nice! I didn't thought it was that simple. THANKYOU

Pi Han Goh - 6 years, 1 month ago

If we do it for $n=3$ only, its pretty simple.

Using our earlier work, the only "new" computation you need is $T_3\left(\ln(x+1)(x+1)^{(x+1)^{x+1}}\right)=x+\frac{x^2}{2}+\frac{5x^3}{6}$ . You realize that this is the same as $T_3\left(\ln(x+1)(x+1)^{x+1}\right)$ ... done!

Thank you so much for the fruitful cooperation! I will use these kinds of problems in the Final Exam in Calculus at Harvard this summer... I will let you know about the results ;)

Otto Bretscher - 6 years, 1 month ago

Yesterday, you suspected that I had a "trick" up my sleeve... well, that was my trick (in the case $n=3$ ) ;)

Otto Bretscher - 6 years, 1 month ago

haha! I thought you had ANOTHER trick up your sleeves! I was persistent in solving it my way but it's over 4 pages long and the equations are enormous so I gave up half way through.

Pi Han Goh - 6 years, 1 month ago

Pi Han Goh please send me a message with an answer to my geometric problem please.

Fawzy Hamdy - 6 years, 1 month ago

@Pi Han Goh so it would be fine right?? If I do the same.

Aditya Kumar - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Prove (or disprove): Let fn[g(x)]f_n[g(x)]fn​[g(x)] denote the nthn^\text{th}nth derivative of g(x)g(x)g(x) at x=1x=1x=1, then fn[n(x)]=fn[n+1(x)]=fn[n+2(x)]=…f_n[^n(x)]=f_n[^{n+1}(x)]=f_n[^{n+2}(x)]=\ldotsfn​[n(x)]=fn​[n+1(x)]=fn​[n+2(x)]=… for all positive integers nnn.

Comments

Prove (or disprove): Let $f_n[g(x)]$ denote the $n^\text{th}$ derivative of $g(x)$ at $x=1$ , then $f_n[^n(x)]=f_n[^{n+1}(x)]=f_n[^{n+2}(x)]=\ldots$ for all positive integers $n$ .