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1 = sin(x) + 2x

And does it imply that an exact answer doesn't even exist? (Or maybe there is an exact answer but my current math skills just don't allow for solving it :)?)

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- Thread starter ehj
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- #1

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1 = sin(x) + 2x

And does it imply that an exact answer doesn't even exist? (Or maybe there is an exact answer but my current math skills just don't allow for solving it :)?)

- #2

daniel_i_l

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If the answer exists, then there's an exact solution (or several). Unfortunately, many times, the exact answer is not known, but when the function is continuous, an approximate answer can be good enough. In the case of the function you're asking about, the solution is usually a transcendental (like pi or e), which is to say that it's not a fraction, and it's not the root to any polynomial. Therefore, there is no common way of writing it other than as the inverse to your function (though you might be able to write it in terms of a complex logarithm, but that doesn't sounds like what you'd want)

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- #5

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try .335418 that should get you pretty damn close :D. (argument in sine is in radians)

- #6

lurflurf

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try

3x-1=0

or

x^3+3x^2-180x+60

for a first approximation.

3x-1=0

or

x^3+3x^2-180x+60

for a first approximation.

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The great mathematician Gauss was known to say that he did not consider it very important to find a symbol to represent the solution to an equation.

Historians have later considered this a kind of "sour grapes" attitude (wrt the results of Galois theory), much like the atittude that Gauss had towards Fermat's last theorem: "I could easily come up with a number of similar propositions that would likewise be exceeedingly difficult to prove; the only interest I have in (FLT) is if it should appear as a corollary of some more general result."

Historians have later considered this a kind of "sour grapes" attitude (wrt the results of Galois theory), much like the atittude that Gauss had towards Fermat's last theorem: "I could easily come up with a number of similar propositions that would likewise be exceeedingly difficult to prove; the only interest I have in (FLT) is if it should appear as a corollary of some more general result."

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- #9

CRGreathouse

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0.3354180323849400594578624117492548397118513151006308650233458056997707969

93488867326823297805099655941760946007417735305449140296885174971232582467743747

67817725202600726615999809586989494215709825435975337739655562116739626518112856

23055310151600910291959720327473494910708608367675240373868492976882512228069498

07934021595150112613380719098738548853419012578533010207808040687825471704745279

60925128393175327764000215850832461166054705141754704841341722301915669142357206

871814814939461336135943871

- #10

Hurkyl

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What you meant to say wasIf the answer exists, then there's an exact solution (or several). Unfortunately, many times, the exact answer is not known

Unfortunately, many times, we want to express the exact answer in a specific form, and it is not known how to do so (or even impossible)

- #11

CRGreathouse

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If you had a function symbol

[tex]\mathcal{Q}(a,b)[/tex] that gave the zero in x (least absolute value, positive if two) for [tex]\sin(x)+ax+b[/tex]

then would you feel better to know that the question has answer [itex]\mathcal{Q}(2,-1)[/itex]?

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If you've got the free time https://www.amazon.com/dp/1402021860/?tag=pfamazon01-20 is a great way to learn about the "unsolvability" of all but the most simple equations.

It should be noted though that methods like newtons method, etc, are often faster at finding answers to third and fourth order problems than exact formulae. With this in mind, using them to find accurate approximate answers to equations should not be viewed as a crippling setback. We often forget that even "exact" answers like [tex]\sqrt{2}[/tex] cannot be represented exactly in an arithmetical calculation, so approximating even the answers to "solvable" equations is something we inevitably have to do anyway.

It should be noted though that methods like newtons method, etc, are often faster at finding answers to third and fourth order problems than exact formulae. With this in mind, using them to find accurate approximate answers to equations should not be viewed as a crippling setback. We often forget that even "exact" answers like [tex]\sqrt{2}[/tex] cannot be represented exactly in an arithmetical calculation, so approximating even the answers to "solvable" equations is something we inevitably have to do anyway.

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- #13

Hurkyl

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Yes it can. In fact, you did so in this every sentence. :tongue: I think what you meant to say isWe often forget that even "exact" answers like [tex]\sqrt{2}[/tex] cannot be represented exactly in an arithmetical calculation,

[itex]\sqrt{2}[/itex] cannot be represented exactly by a terminating decimal number

- #14

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...

If you had a function symbol

[tex]\mathcal{Q}(a,b)[/tex] that gave the zero in x (least absolute value, positive if two) for [tex]\sin(x)+ax+b[/tex]

then would you feel better to know that the question has answer [itex]\mathcal{Q}(2,-1)[/itex]?

I find this to be a clever response

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If you had a function symbol

[tex]\mathcal{Q}(a,b)[/tex] that gave the zero in x (least absolute value, positive if two) for [tex]\sin(x)+ax+b[/tex]

then would you feel better to know that the question has answer [itex]\mathcal{Q}(2,-1)[/itex]?

Yes, in fact it makes me feel smug and superior that I can go into a room and be confident that most likely only I have enough mathematical training to know what the Q function is.

- #16

CRGreathouse

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Yes, in fact it makes me feel smug and superior that I can go into a room and be confident that most likely only I have enough mathematical training to know what the Q function is.

Ooh, this tempts me to write an article for the PF Library defining the Q function so your post gets auto-linked.

- #17

mathwonk

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for that matter what do you consider a number? i.e., is it a finite decimal? a sequence of approximations? so would a newton's process for approximating an answer satisfy you?

if so, then one can solve most equations. if not, then one cannot even solve the one above.

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- #19

mathwonk

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and i have already alluded to the process of using newtons approximations, which "solve" a wide variety of equations.

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[tex]\sum_{k=0}^{\infty} f(k)[/tex]

- #21

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[tex]\sum_{k=0}^{\infty} f(k)[/tex]

Any solution to an equation of the type you are asking about is a real (or maybe complex) number, and every real (and complex) number can be written in that form because the real and complex numbers are complete

- #22

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Yes I can see that, but is it always possible toAny solution to an equation of the type you are asking about is a real (or maybe complex) number, and every real (and complex) number can be written in that form because the real and complex numbers are complete

- #23

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I find this to be a clever response

Agree very much in this. Maybe it was this Hermann Weyl meant by "We are left with

our symbols" - although he of course meant symbols representing more advanced things.

By using symbols everything can be solved exactly and explicitely. Define a symbol for the problem and the problem is solved using this symbol. Maybe even integers can be regarded as nondefined symbols in some way? I.e a numerical exact answer may be regarded as fundamentally resting on the integer symbols (1, 2, 3, ...) contained?

Symbols maybe the true clue to progress of mankind.

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- #24

CRGreathouse

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But if we allow the use of an infinite sum of elementary functions, is it then always possible to solve any equation and get an exact answer?

Sure, in the sense that you can express any real number as its decimal expansion.

If the equation uses only elementary functions, then you could even encode (presumably) the bisection formula.

- #25

mathwonk

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if what you say were true, it should be possible to say whether the sum of a series were say rational or not. so if you believe the values of the zeta function are exact expressions, are they rational?

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If you had a function symbol

[tex]\mathcal{Q}(a,b)[/tex] that gave the zero in x (least absolute value, positive if two) for [tex]\sin(x)+ax+b[/tex]

then would you feel better to know that the question has answer [itex]\mathcal{Q}(2,-1)[/itex]?

Since I am by far the dumbest person the post in this thread, can someone explain this to me, because it seems to be very smart-aleck

- #27

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Since I am by far the dumbest person the post in this thread

ok

can someone explain this to me, because it seems to be very smart-aleck

Well the original question boils down to how we can represent numbers that are exact solutions to certain equations. The number representation and usual definitions of operators we have may not be a nice way to show the solution. (ie (sqrt(5) - 1)/pi).

We can write out certain solutions to a point, 5.293782093... etc but we might not be able to represent these irrational numbers with the usual operators and symbols that we have. However, for any one equation we can always define a new operator, or symbol that represents exactly that! (the solution to our equation).

I don't think it was smart aleck, but an intellegent clever response, when you really take it all in.

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And thanks a lot for clearing that up

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- #30

arildno

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1. Is the equation soluble? (I.e, does there exist a solution? This can often be resolved without specifying that solution!)

2. Is an solution of an equation unique, or are there other solutions as well?

3. How would the solution (or solutions) change if we were to tweak the equation a bit?

(I.e, what is the "behaviour" of solutions to a close-knit family of equations?)

4. Can we devise an algorithm that will guarantee us to get an arbitrarily good approximation to the solution of the equation?

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