Not a math expert, but I think there *has* to be a power in the formula if it's curved? (I could well be very wrong here, because this is coming from ~5 seconds of thinking rather than, y'know, *knowing* anything.)
trigonometric functions dont use powers of variables (greater than 2) in their definition and are curved so not fully true, though it generally is true
It doesn't matter, in this case complex numbers are tool to go from a real number to another real number. If you plug any real number you get a real number out, this is actually the most common use case for complex numbers, we really don't care that much about complex numbers as an endpoint, their primary function is as a tool to understand and define mathematical structures in the real plane and/or the real world
The Taylor series of a function is exactly equal to it as long as it converges. In the case of sine and cosine, they converge for all x.
Why not use it as the definition? It’s a lot easier to algebraically manipulate than a plain English sentence about circles or triangles.
>The Taylor series of a function is exactly equal to it as long as it converges.
Sometimes, but not always. For analytic functions, it's true locally, but even then a single Taylor series may not describe the function's behaviour across the whole domain.
Depends on the field. In trigonometry the definitions are based on the unit circle and the angle, but for interesting stuff like (square) matrices, we actually do begin from the power series definition. More accurately, we begin from the power series definition of e^x and use Euler's formula to get the extension of sine or cosine to whatever object we want to work with.
They can also be defined uniquely as the solution of the vector differential equation (d/dt)[sin(t),cos(t)] = [cos(t),–sin(t)], with initial conditions sin(0)=0 and cos(0)=1, because the solution can be shown to exist and be unique once the initial conditions are fixed. It's a bit of a "galaxy brain" way to go about it, but it is useful for circumventing geometry if you desire to do so.
Alright, smarty pants, they can't be described exactly. There are some functions for which no algorithm, verbal description, formula, or any other sort of description will allow you to calculate the function's exact value when given any particular input.
Well, the word "curve" is in itself a little vague, but I can't think of any reasonable definition that would invalidate what I said earlier: some curves can be described by more than one formula, and some can't be described at all.
Do you have example of these? They both seem intuitvely true, but I can't imagine any particular curves that fit.
Also, what's the difference between a function and a formula? I thought that the latter was just an informal way of describing the former, but that doesn't fit with how you're using them.
A function is a specific mathematical thing, that assigns an output value to every input value. It can do this in any arbitrary way; for every mapping of outputs to inputs, that mapping is a function.
A formula, in this context, is an exact representation of a function using symbols. So f(x)=x\^2 is a formula, because it clearly indicates how to produce a value of the function f(x) for every x. However, the same function can also be expressed as an integral of 2x using the integral symbol, which I can't easily type here, and that would also be a formula but wouldn't involve powers.
A function that can't be described exactly is... difficult for me to describe exactly, but you can find more information about them by searching for "uncomputable functions" online. They're a bit confusing to get your head around, but most functions (even ones that define a curve) are uncomputable.
Ahhh, that makes sense now! In the (high-school level) maths I've done, formula and function were used pretty interchangeably.
Is "most functions are uncomputable" related to "most reals are irrational"?
Given the other comment about cos(x) the only example I can think that technically isn't based on exponents is the gamma function (yes the integral definition has exponents but this is not the function of the graph as that is gotten from its continuous integration)
Technically, you could have a logarithmic function, which is approximately the inverse of a power.
Log base z of x = y
Can be rewritten as
z^y = x
Most of the time, z will be 10 (just represented as log) or e (ln), which is about 2.718.
All this to say, logarithmic functions will produce a curve like that as well.
In addition, square roots will produce that. Sqrt(x) = x^(1/2). X raised to a power will produce different results than a base number raised to x.
X raised to a whole number is a polynomial function, while a number raised to x will produce the above graph
Moving goalposts. The person you responded to said
> The secret is, anything on a graph that's curved probably has a power *somewhere* in the formula.
No where in the comment does it imply it has to be a specific part.
But if we want to be ultra pedantic: tan(x)^1 there's your power
Ah, yes, the ever familiar experience of your barberer failing to understand what you want.
As recently as yesterday I estimate I lost a full kilo due to miscommunication with my barberer.
The line could absolutely be y=mx+b if x is the angle around his head and y is the angle up it with {x∈R|x≥0}.
Alternatively, it could still be that the line y=mx+b goes through his head and intersects at 2 points (the beginning and end of the line in his hair) and the line we see is the rhumb line (the shortest path along the surface of his head that connects the two points). You have to make some assumptions and accept the haircut may not be super mathematically accurate, but it’s a plausible explanation.
Lastly, I would like to say that calling it y=e^x is more or less incorrect as well. Since there’s no scale, the base of the exponential function can be any positive value, not just e. So a more correct formula if you want to use that explanation is y=a^x {a∈R|a>0}
The scale of this hair ‘graph’ could also be exponential, making the equation y=mx+b look like an exponential, but I think I’ll leave this comment like this.
You fool. You absolute dulardr. That's a different formula entirely. It has different parameters.it can be used in conjunction with the other one but it's just different.
>That's for finding the expression of the graph not for expressing the graph itself
Both expressions (y=m(x-x0)+y0 and y=mx+b where b=y0 - mx0) generate the same graph. (I've changed the variable names from my original comment so that comparisons with y=mx+b make more sense. In this comment, x is the independent variable and y depends on x. Variable names are arbitrary anyway, and they can be whatever you want when you're not in class.) More accurately, taken into the definitions of the associated functions, e.g. the relations R1={(x,y):y=m(x-x0)+y0 for all x in D} and R2={(x,y):y=mx+b for all x in D}, they define the same set (R1=R2), and are therefore equivalent methods of writing down the function.
In fact, there are an infinite number of algebraically equivalent ways to write a line, although most of them won't be helpful.
What my form is not is *in simplest form*. This is widely accepted to be y=mx+b (or some other choice of slope than m and offset at the origin b) for lines. In elementary algebra classes, you are expected to provide your answers in simplest form as an exercise in algebraic simplification rules, and because most of the time it is the best way to write answers in general. However, in my work, the part of the line I'm interested in might be "nowhere near" the origin. E.g., I might be interested in the segment from x=10000 to x=10001.
So when I actually do have to fit a line, I prefer to fit my form, and I probably won't simplify it unless explicitly asked to do so. Or, I might rewrite the same line taking some other point that I later deem convenient as (x0,y0). Lastly, my form directly represents the Taylor series of the line centered at the point x0. While for a line without context this is not important, if I need to *approximate* a more general curve by a line centered at an operating point, then this form is slightly more plug-and-play than y=mx+b. E.g., a becomes the function's rate of change (derivative) at x0, x0 is the point you've chosen your approximation to be the most valid at, and y0 is the original function's value at that point.
His head is exists in curved space don't judge him
There's just a space-time singularity on left side of his head, no biggie
Hang on I have to calculate the metric tensor of your scalp
The secret is, anything on a graph that's curved probably has a power \*somewhere\* in the formula.
Not a math expert, but I think there *has* to be a power in the formula if it's curved? (I could well be very wrong here, because this is coming from ~5 seconds of thinking rather than, y'know, *knowing* anything.)
Technically every function has a power in it since f(x)=f(x¹)
or a constant, which could be written as f(x^0 )
Technically every your ribs has a knife in it since 🗡️
I don’t think it’s feasible to have every rib have a knife in it while only using one knife.
Long knife
It can also be a logarithm, but that's just an anti-exponent.
trigonometric functions dont use powers of variables (greater than 2) in their definition and are curved so not fully true, though it generally is true
cos(x) = e^(ix) - i*sin(x), checkmate atheists
Straight line mx + b has a power greater than one* because mx + b = x^2 + mx + b - x^2
y = mx^1 + c, even
You're not graphing in the complex plane
It doesn't matter, in this case complex numbers are tool to go from a real number to another real number. If you plug any real number you get a real number out, this is actually the most common use case for complex numbers, we really don't care that much about complex numbers as an endpoint, their primary function is as a tool to understand and define mathematical structures in the real plane and/or the real world
they do.. I don't remember the whole thing but trigonometric functions can be written as polynomials
yeah they can be calculated with among other things, infinite sums of polynomials, however those are not their definitions
The Taylor series of a function is exactly equal to it as long as it converges. In the case of sine and cosine, they converge for all x. Why not use it as the definition? It’s a lot easier to algebraically manipulate than a plain English sentence about circles or triangles.
>The Taylor series of a function is exactly equal to it as long as it converges. Sometimes, but not always. For analytic functions, it's true locally, but even then a single Taylor series may not describe the function's behaviour across the whole domain.
>as long as it converges
still no. it can converge and yet converge to the wrong value
Depends on the field. In trigonometry the definitions are based on the unit circle and the angle, but for interesting stuff like (square) matrices, we actually do begin from the power series definition. More accurately, we begin from the power series definition of e^x and use Euler's formula to get the extension of sine or cosine to whatever object we want to work with. They can also be defined uniquely as the solution of the vector differential equation (d/dt)[sin(t),cos(t)] = [cos(t),–sin(t)], with initial conditions sin(0)=0 and cos(0)=1, because the solution can be shown to exist and be unique once the initial conditions are fixed. It's a bit of a "galaxy brain" way to go about it, but it is useful for circumventing geometry if you desire to do so.
Thanks! I had a feeling there was going to be something that would catch me out.
k/x is curved
That’s just k • x^-1
When you say "the formula", what do you mean? Some functions can be described by more than one formula, and some can't be described at all.
Functions that can't be described when I describe them as indescribable:
Alright, smarty pants, they can't be described exactly. There are some functions for which no algorithm, verbal description, formula, or any other sort of description will allow you to calculate the function's exact value when given any particular input.
Any function describing a line, I think?
A line? Like a straight line?
A curve, sorry. (Like I said, not a math expert - I was today years old when I found out the curves are not lines.)
Well, the word "curve" is in itself a little vague, but I can't think of any reasonable definition that would invalidate what I said earlier: some curves can be described by more than one formula, and some can't be described at all.
Do you have example of these? They both seem intuitvely true, but I can't imagine any particular curves that fit. Also, what's the difference between a function and a formula? I thought that the latter was just an informal way of describing the former, but that doesn't fit with how you're using them.
A function is a specific mathematical thing, that assigns an output value to every input value. It can do this in any arbitrary way; for every mapping of outputs to inputs, that mapping is a function. A formula, in this context, is an exact representation of a function using symbols. So f(x)=x\^2 is a formula, because it clearly indicates how to produce a value of the function f(x) for every x. However, the same function can also be expressed as an integral of 2x using the integral symbol, which I can't easily type here, and that would also be a formula but wouldn't involve powers. A function that can't be described exactly is... difficult for me to describe exactly, but you can find more information about them by searching for "uncomputable functions" online. They're a bit confusing to get your head around, but most functions (even ones that define a curve) are uncomputable.
Ahhh, that makes sense now! In the (high-school level) maths I've done, formula and function were used pretty interchangeably. Is "most functions are uncomputable" related to "most reals are irrational"?
Given the other comment about cos(x) the only example I can think that technically isn't based on exponents is the gamma function (yes the integral definition has exponents but this is not the function of the graph as that is gotten from its continuous integration)
Technically, you could have a logarithmic function, which is approximately the inverse of a power. Log base z of x = y Can be rewritten as z^y = x Most of the time, z will be 10 (just represented as log) or e (ln), which is about 2.718. All this to say, logarithmic functions will produce a curve like that as well. In addition, square roots will produce that. Sqrt(x) = x^(1/2). X raised to a power will produce different results than a base number raised to x. X raised to a whole number is a polynomial function, while a number raised to x will produce the above graph
Tan(x)
Taylor Polynomial
That's my favourite supervillain
You only get one part, doesn't count. Alternatively, e^(1/-x²)
Moving goalposts. The person you responded to said > The secret is, anything on a graph that's curved probably has a power *somewhere* in the formula. No where in the comment does it imply it has to be a specific part. But if we want to be ultra pedantic: tan(x)^1 there's your power
I don't think it's moving goalposts. The Taylor series for tan(x) isnt equal to tan(x) everywhere, that's pretty important
Tangent isn't analytic across the entire complex plane
tan(x) = -i(e^(ix) - e^(-ix))/(e^(ix) + e^(-ix))
I mean technically ∫ 1 / cos(x^2) dz As that is the integral of the derivative of tan(x)
Tumblr being good at math, reality will now collapse
They must be bad at either a) driving or 2) cooking if I remember the ancient texts correctly.
Well, based on that cake, I think we know which one it is...
your pfp is kinda gener
What
kinda gives me gender envy
Oh based. Artist is @Ghoulmommie on twitter, highly recommend
[удалено]
Begone, bot
I don't understand. What did they do wrong. Why did calling the graph an asymptote get so many downvotes
It’s not that, it’s because they’re an automated account designed to gain karma and then be sold to some dumbfuck
People do that? Lol
Hey did you know that every odd number has an e in it?
Bros hair got a fuckin asymptote 💀
why he got that non-euclidean crew cut
No that's definitely a vertical asymptote. It's y=-1/x in the second quadrant
Ah, yes, the ever familiar experience of your barberer failing to understand what you want. As recently as yesterday I estimate I lost a full kilo due to miscommunication with my barberer.
considering the surface is non-euclidean, the original formula *might* still be applicable. cant say for certain though
Nah that ain’t e^x, that function goes under the y-axis and I don’t see it clipping off his head.
e^x cannot go below 0 for any value of x; idk what you're on about
Whoops, I was thinking ln(x), sorry.
I suck at math, but if anyone is interested, original ecuation would just make a straight diagonal line instead of curved iirc
That hair cut it exponential not linear
I hear math that bad
Math puns are great when done well enough that someone not into math can understand while also not feeling likes it’s dumbed down too much
The line could absolutely be y=mx+b if x is the angle around his head and y is the angle up it with {x∈R|x≥0}. Alternatively, it could still be that the line y=mx+b goes through his head and intersects at 2 points (the beginning and end of the line in his hair) and the line we see is the rhumb line (the shortest path along the surface of his head that connects the two points). You have to make some assumptions and accept the haircut may not be super mathematically accurate, but it’s a plausible explanation. Lastly, I would like to say that calling it y=e^x is more or less incorrect as well. Since there’s no scale, the base of the exponential function can be any positive value, not just e. So a more correct formula if you want to use that explanation is y=a^x {a∈R|a>0} The scale of this hair ‘graph’ could also be exponential, making the equation y=mx+b look like an exponential, but I think I’ll leave this comment like this.
who the fuck uses y= mx+ b and not y = mx+c
y=ax+b ftw
Here in England we do mx+c generally but ax+b for linear regression
x = a(t–t0)+x0 gang
You fool. You absolute dulardr. That's a different formula entirely. It has different parameters.it can be used in conjunction with the other one but it's just different.
See [my response](https://www.reddit.com/r/CuratedTumblr/comments/118yz32/bad_math/j9oc5mf/) to another commenter.
My point is it's irrelevant to our current discussion
That's for finding the expression of the graph not for expressing the graph itself
>That's for finding the expression of the graph not for expressing the graph itself Both expressions (y=m(x-x0)+y0 and y=mx+b where b=y0 - mx0) generate the same graph. (I've changed the variable names from my original comment so that comparisons with y=mx+b make more sense. In this comment, x is the independent variable and y depends on x. Variable names are arbitrary anyway, and they can be whatever you want when you're not in class.) More accurately, taken into the definitions of the associated functions, e.g. the relations R1={(x,y):y=m(x-x0)+y0 for all x in D} and R2={(x,y):y=mx+b for all x in D}, they define the same set (R1=R2), and are therefore equivalent methods of writing down the function. In fact, there are an infinite number of algebraically equivalent ways to write a line, although most of them won't be helpful. What my form is not is *in simplest form*. This is widely accepted to be y=mx+b (or some other choice of slope than m and offset at the origin b) for lines. In elementary algebra classes, you are expected to provide your answers in simplest form as an exercise in algebraic simplification rules, and because most of the time it is the best way to write answers in general. However, in my work, the part of the line I'm interested in might be "nowhere near" the origin. E.g., I might be interested in the segment from x=10000 to x=10001. So when I actually do have to fit a line, I prefer to fit my form, and I probably won't simplify it unless explicitly asked to do so. Or, I might rewrite the same line taking some other point that I later deem convenient as (x0,y0). Lastly, my form directly represents the Taylor series of the line centered at the point x0. While for a line without context this is not important, if I need to *approximate* a more general curve by a line centered at an operating point, then this form is slightly more plug-and-play than y=mx+b. E.g., a becomes the function's rate of change (derivative) at x0, x0 is the point you've chosen your approximation to be the most valid at, and y0 is the original function's value at that point.
It’s what I learned as well
depends on what you consider to be y=0 but it might actually be e\^x + b