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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 \(n\).

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Note by Pi Han Goh
1 year, 8 months ago

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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 · 1 year, 8 months ago

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@Otto Bretscher 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 · 1 year, 8 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 · 1 year, 8 months ago

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@Otto Bretscher 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 · 1 year, 8 months ago

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@Pi Han Goh 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 · 1 year, 8 months ago

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@Pi Han Goh so it would be fine right?? If I do the same. Aditya Kumar · 1 year, 5 months ago

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Pi Han Goh please send me a message with an answer to my geometric problem please. Fawzy Hamdy · 1 year, 8 months ago

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