# Daedalus' Fourth Challenge

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My senior year in high school, I discovered a way to manipulate exponential functions to produce any sequence of numbers I desired. The challenge I am setting forth asks for the exponential function that produces the sequence:

Edit:

{{1,1}, {2,2}, {3,3}, {4,4}, {5,5}}

That is, when $x$ is equal to one, $y$ is equal to one and so forth up to {5, 5}.

{1, 2, 3, 4, 5}

Such that:

$F(1)=1$

$F(2)=2$

$F(3)=3$

$F(4)=4$

$F(5)=5$

The answer I am looking for only uses exponential functions of the form $a^{f(x)} \, b^{g(x)} \, c^{h(x)} \text{etc...}$ You are only allowed to use elementary operators. No infinite series, trigonometry, or any mathematics higher than algebra. The result must be in the form as expressed above and contain only one term.

I will post the answer along with the formula in a couple of weeks unless someone can solve this problem : )

Edited by Daedalus

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Er, I'm not sure what you're asking. Are you saying you just want y set equal to some base raised to a power including x, such that that sequence is produced?

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Hi John. I haven't had much sleep lately. I just realized that the sequence of points wasn't properly structured in a way that makes sense. I've been working with lists in Mathematica and I sometimes mix things up. Basically, what I am asking for is an exponential function that produces the following:

$F(1)=1$

$F(2)=2$

$F(3)=3$

$F(4)=4$

$F(5)=5$

The function must be in the form:

$F(x)=a^{f(x)} \, b^{g(x)} \, c^{h(x)}$ etc...

Where $a$, $b$, $c$, etc.. are constants and $f(x)$, $g(x)$, $h(x)$, etc.. are functions of $x$ such as $x^2+2\, x + 1$.

I designed the challenge to show how an exponential function can approximate a linear function within a closed interval such as $\{x : 1 \le x \le 5\}$ and to demonstrate how we can derive an exponential function for any sequence of numbers.

I'll give you a hint in that the solution involves mathematics similiar to the answer I gave for the second part of Daedalus' Second Challenge. Except I am only looking for the exponential function which produces the above sequence and not the formula that I derived to solve it : ) However, I'll give a five star rating to anyone who can provide the formula along with the exponential function.

Edited by Daedalus

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Since no one has provided me with the answer, I will go ahead and post the solution:

The following exponential function is the solution to the challenge:

$2^{-\frac{(x-5)(x-3)(x-1)(3x-8)}{6}} \times 3^{\frac{(x-5)(x-4)(x-2)(x-1)}{4}} \times 5^{\frac{(x-4)(x-3)(x-2)(x-1)}{24}}$

The function produces the following graph:

We can see that the function approximates $y=x$ on the interval, $\{x \in \mathbb{R} \, |\, 1 \le x \le 5\}$.

The Process:

I figured out how to do this when I discovered Newton's interpolation formula my 11th grade year in high school. I didn't know about interpolation at the time. I was in Trigonometry and had learned that summations of $x$ had a polynomial that generalized the sum. It took me one year and three months to discover the equation which predicted the summations of $x^p$. I did not have a computer and I did all of the calculation using paper and a TI-88 (I sure do love Mathematica... It has saved a lot of trees):

$F(s,\, n,\, p)=\sum_{j_{1}=1}^n \sum_{j_{2}=1}^{j_{1}} ... \sum_{x=1}^{j_{S}} (x^p)=\sum_{j=0}^p \left((-1)^{j+p} \left(\sum_{k=0}^j \frac{k^p \, \left \langle -j \right \rangle_{j-k}}{(j-k)!}\right)\left(\frac{\left \langle n \right \rangle_{j+s}}{(j+s)!} \right)\right)$

Where $s$ represents the recursion level of the summation (when $s=0$ we get the original sequence and when $s=1$ we get the first summation of the sequence and so on), and we sum the function, $x^p$, from $0$ to $n$ where $p$ is the power.

The process, which involved recursively taking the deltas of the number sequences and applying the rising factorial or Pochhammer function, allowed me to derive the exponential version by recursively dividing instead of doing a subtraction. This is demonstrated below (note: I replaced the Pochhammer function with a product series operator for those that are not familiar with the rising factorial):

Newton's Interpolation:

Recursively take the deltas of each sequence:

$\begin{matrix} & & & y_4 - 3\, y_3 + 3\, y_2 - y_1\\ & & y_3 - 2\, y_2 + y_1 & y_4 - 2\, y_3 + y_2\\ & y_2 - y_1 & y_3 - y_2 & y_4 - y_3 \\ y_1 & y_2 & y_3 & y_4 \end{matrix}$

We are only interested in the results at the top of each column. Also, we can see that each result is an alternating sum of the original sequence with coefficients that are pascal numbers. Next, we use each result with the rising factorial and standard factorial as follows:

$\left(y_1\right)\frac{1}{0!}\ -\ \left(y_2 - y_1\right)\frac{(1-x)}{1!}\ +\ \left(y_3 - 2\, y_2 + y_1\right)\frac{(1-x)(2-x)}{2!}\ -\ \left(y_4 - 3\, y_3 + 3\, y_2 - y_1\right)\frac{(1-x)(2-x)(3-x)}{3!}\ +\ \text{etc...}$

The formula which defines this process is as defined below (note: I have expanded the formula to include recursively summing the sequence such that when $s=0$ we get the original sequence and when $s=1$ we get the first summation of the sequence and so on):

$F(x, \, s,\, n) = \sum_{i=0}^{n-1}\left( \sum_{j=0}^{i}\left(f(j)\frac{(-1)^{j}\ i!}{j!\ (i-j)!}\right) \frac{(-1)^{s}}{(i+s)!} \prod_{k=1}^{i+s}\left(k-s-x\right)\right)$

Where $x$ is the variable, $s$ is the summation as explained above, and we sum from $0$ to $n-1$. We can start from any number by modifying $f(j)$ to include a starting index, $f(j+start)$.

Daedalus' Exponential Interpolation:

We must do the same thing as above except we will divide the numbers of the sequence:

$\begin{matrix} & & & y_4^1 \times y_3^{-3} \times y_2^{3} \times y_1^{-1}\\ & & y_3^1 \times y_2^{-2} \times y_1^1 & y_4^1 \times y_3^{-2} \times y_2^1\\ & y_2^1 \times y_1^{-1} & y_3^1 \times y_2^{-1} & y_4^1 \times y_3^{-1} \\ y_1 & y_2 & y_3 & y_4 \end{matrix}$

The next part is also similar except additions become multiplications, subtractions become division, and multiplication becomes exponents:

$\left(y_1\right)^{\frac{1}{0!}}\ \times\ \left(y_2^1 \times y_1^{-1}\right)^{-\frac{(1-x)}{1!}}\ \times \ \left(y_3^1 \times y_2^{-2} \times y_1^1\right)^{\frac{(1-x)(2-x)}{2!}}\ \times \ \left(y_4^1 \times y_3^{-3} \times y_2^{3} \times y_1^{-1}\right)^{-\frac{(1-x)(2-x)(3-x)}{3!}}\ \times \ \text{etc...}$

The following formula defines the above process such that when $p=0$ we get the original sequence and when $p=1$ we get the first product of the sequence and so on:

$F(x, \, p,\, n) = \prod_{i=0}^{n-1}\left( \prod_{j=0}^{i}\left(f(j)^{\frac{(-1)^{j}\, i!}{j!\, (i-j)!}}\right)^{ \frac{(-1)^{p}}{(i+p)!} \displaystyle \prod_{k=1}^{i+p}\left(k-p-x\right)}\right)$

Where $x$ is the variable, $p$ is the recursion level of the product as explained above, and we multiply the outputs from $0$ to $n-1$. We can start from any number by modifying $f(j)$ to include a starting index, $f(j+start)$.

This method can be applied to any operator that obeys the associative law (which is why it only works for summations and products). Please forgive any grammar errors in this post. I broke a tooth over thanksgiving and it is causing me a tremendous amount of pain. It's 3:00 am and I have a dentist appointment today. So... I kinda rushed this post. I will correct any issues tomorrow as long as the edit timer has not expired.

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