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The city had a new transmogrifier. I was willing to go in, but it would be boring, I was told. Meanwhile, some people were bragging about how they would come out the transmogrifier.

One said, "My little buddy should jump down, but there will be a new one."

Jumpy said, "Why why, if I go in it 4 times I come back!"

His brother said, "I do too!"

Jumpy said, "But you will have a negative, and I won't!"

Jumpy's brother said, "Oh, oh, oh, but you will become me!"

The Famous one argued, "You all are shameful. I go back to myself!"

His family said, "Well, I'm lucky too, we're really close, but we have an ugly multiplier next to us!"

The Famous one said, "I don't! That's one!"

I went back to myself too if I went in, but people didn't care about me.

What is the transmogrifier, and what am I?

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I believe the Transmogrifier is:

the differentiation process in mathematics.

Let's see this by examining what it does to some of the characters in the story...

One said, "My little buddy should jump down, but there will be a new one."

The character speaking here is the algebraic function $x^a$. When differentiated, this becomes $ax^{a-1}$ - his 'little buddy' is his exponent, who 'jumps down' but is replaced in the exponent by a value one lower than before.

Jumpy said, "Why why, if I go in it 4 times I come back!"
His brother said, "I do too!"

'Jumpy', here, is a separate function: $sin(x)$. His name is Jumpy because of the nature of the graph of $y=sin(x)$, permanently oscillating (jumping) between extreme values of 1 and -1.

Applying differentiation repeatedly to this function yields a cycle of:
$sin(x)$ --> $cos(x)$ --> $-sin(x)$ --> $-cos(x)$ --> $sin(x)$ (etc.)

In other words, if you apply differentiation 4 times, he returns to his original form. This is also the case for Jumpy's brother, who is the related function $cos(x)$ - starting one step further into the cycle as displayed above, likewise he also ends up returning to his original form after 4 iterations.

Jumpy said, "But you will have a negative, and I won't!"
Jumpy's brother said, "Oh, oh, oh, but you will become me!"

Their further comments concern what happens to them after just one visit to the Transmogrifier: Jumpy's brother, $cos(x)$, gains a negative sign as part of its transformation, while Jumpy ($sin(x)$) simply becomes his brother, $cos(x)$.

The Famous one argued, "You all are shameful. I go back to myself!"
His family said, "Well, I'm lucky too, we're really close, but we have an ugly multiplier next to us!"
The Famous one said, "I don't! That's one!"

'The Famous One' is the exponential function, $e^x$, which when differentiated remains as $e^x$. Its family are the set of exponential functions which have a multiplier in the exponent, of the form $e^{ax}$ - these change only slightly with differentiation, becoming $ae^{ax}$ afterwards, and if differentiated repeatedly the multiplier by their side becomes uglier and uglier (increasing powers of $a$). Really, The Famous One is just a special case of this family where the value of this multiplier is simply 1 (explaining his comment, "That's one!").

I went back to myself too if I went in, but people didn't care about me.

There is one other function which remains unchanged upon differentiation - the function where $y = 0$. This must be our unnamed narrator. Since this is really a trivial case, people 'don't care about him'!

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  • $\begingroup$ @msh210 I agree, and that multiplier would continue getting uglier after each round. I'll make that minor adjustment, thanks :) $\endgroup$ – Stiv Dec 27 '20 at 22:36
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    $\begingroup$ Bit of a problem with Jumpy's statement "But you will have a negative, and I won't". Where it is in the conversation, it would refer to the brother's claim to come back after 4 times. But that would be false. The brother comes back exactly the same after 4 transmorgrifications, just like Jumpy. No negatives. Where the negative sign comes in is in turning into each other, which is only discussed after Jumpy's remark. $\endgroup$ – Paul Sinclair Dec 28 '20 at 18:10
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    $\begingroup$ @PaulSinclair Sorry if the riddle was a bit unclear, but what I meant was that that statement was after only one visit. $\endgroup$ – new QOpenGLWidget Dec 29 '20 at 1:05
  • $\begingroup$ @zixuanisbadatPuzzling - it happens. If I were really concerned by it, I would have said more. The evidence that Stiv had the correct answer was overwhelming, despite requiring a twisted meaning for one phrase. But try always to consider your puzzle from the readers' point of view, remembering that they can only see what you write, not what you mean by it. $\endgroup$ – Paul Sinclair Dec 29 '20 at 1:16
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I see the green checkmark next to the other answer, but this fits way too well to ignore:

The transmogrifier is the "multiply by yourself" operation.

One said, "My little buddy should jump down, but there will be a new one."

Taking the name literally, "One" is 1, and his little buddy is -1. Were -1 to jump into the transmogrifier, you'd multiply -1 by itself, and there would be a new 1.

Jumpy said, "Why why, if I go in it 4 times I come back!"

Jumpy is the imaginary unit $i$. If he goes in four times, we get $i\times i\times i\times i\times i= i$, where the value jumps around the unit circle as the multiplications progress.

His brother said, "I do too!"

Jumpy's brother is $-i$, and $-i\times -i\times -i\times -i\times -i = -i$.

Jumpy said, "But you will have a negative, and I won't!"

$-i$ does, indeed, have a minus sign

Jumpy's brother said, "Oh, oh, oh, but you will become me!"

In the process of going around the unit circle and becoming themselves again, the brothers $i$ and $-i$ indeed temporarily become one another.

The Famous one argued, "You all are shameful. I go back to myself!"

That would be again the famous "one", the only member of the family (see below) that instantly becomes itself when multiplied by itself

His family said, "Well, I'm lucky too, we're really close, but we have an ugly multiplier next to us!"

The family is "roots of unity", where the four aforementioned brothers (1,-1,i,and-i) don't have any ugly multipliers preceding their real and imaginary parts, but the infinitely many other roots of unity do. They all live densely packed on the unit circle.

The Famous one said, "I don't! That's one!"

"one" seems have an inflated ego; While it's true that he doesn't have a multiplier, he seems to be forgetting that as a member of an infinite set, he makes up for exactly 0% of all the members. Maybe the fame of being the most famous number of all has gone to his head.

I went back to myself too if I went in, but people didn't care about me.

You were even missing from many early number systems, because people didn't care about you

What is the transmogrifier, and what am I?

The transmogrifier is the "multiply by yourself" operation, and you are the number zero.

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  • $\begingroup$ Clue 9 doesn't really fit well, but all the others do too. Still a +1. $\endgroup$ – new QOpenGLWidget Dec 27 '20 at 23:17
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    $\begingroup$ This also doesn't explain the name "Jumpy", but everything else does fit well. $\endgroup$ – bobble Dec 27 '20 at 23:25
  • $\begingroup$ Also, rot13(Whzcl naq uvf oebgure qba'g ernyyl svg - gurl jbhyq obgu orpbzr bar (fvapr gur bcrengvba vf "envfr gb gur cbjre bs gjb", abg "zhygvcyl ol jung lbh bevtvanyyl jrer")) $\endgroup$ – Mariia Mykhailova Dec 29 '20 at 0:40

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