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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/

Hey Abhishek, we think your problem is great for the following reason:

  • Astonishing result: It is interesting to find that even after having equal top speeds, Tom will never be able to catch Jerry.

We have made the following key improvement to your problem that makes it even better:

  • Imagery: Added a skin to the problem. Adding the story of Tom chasing Jerry makes the problem setup more interesting and engaging.

Note by Pi Han Goh
2 years, 7 months ago

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1 vote

  Easy Math Editor

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

Otto Bretscher - 2 years, 7 months ago

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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 - 2 years, 7 months ago

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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 - 2 years, 7 months ago

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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 - 2 years, 7 months ago

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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 - 2 years, 7 months ago

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

Aditya Kumar - 2 years, 4 months ago

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

Fawzy Hamdy - 2 years, 7 months ago

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