Evaluating a Function | Challenging Yet Rewarding
Вставка
- Опубліковано 19 жов 2023
- 🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
/ @aplusbi
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/stores/sybermat...
Follow me → / sybermath
Subscribe → ua-cam.com/users/SyberMath?sub...
⭐ Suggest → forms.gle/A5bGhTyZqYw937W58
If you need to post a picture of your solution or idea:
intent/tweet?text...
#functions #functionalequations
via @UA-cam @Apple @Desmos @NotabilityApp @googledocs @canva
An Exponential Equation | 2^{3^{5^x}} = 3^{2^{5^x}}: • An Exponential Equatio...
A Surprising Exponential Equation: • A Surprising Exponenti...
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus
My first step was finding f(0) = -5; then f(2) = 1+1+5=7;f(3)=1+7+5=13. Incrementing x by 1 increments the function by 6; f(x) = 6x - 5; plug in -17/6 --> -17-5 = -22.
This I did too.
I did something like this too, except I also went through the effort of proving it for negative numbers and rational numbers. Following that pattern only really proves it for the integers >= 0.
I wonder if it is possible to prove this formula for all real numbers.
the result is right but the approach is wrong. see the function in not defined specifically for integer domain.. so your recurring pattern is just being verified with integers only. if you could do it like f(x+dx)=f(x)+6 where dx is infinitesimally small real number.. then its right. now if you take limits y->0.. then it is same as dx. now lim y->0 [f(x+y)-f(x)]/y=lim y->0 [(f(y)+5)/6]=f'(0).. and the rest i did it in my comment you may check it out.
@@rosettaroberts8053 It is possible to prove this formula true for all real numbers, if f is continuous. I am not sure if it's true for non-continuous, but I think it is.
@@chaosredefined3834 If you look up Cauchy's functional equation, you'll find some discontinuous examples. It turns out that all the nonlinear solutions are really awful, though - they're unbounded on any interval, for example.
It's actually quite easy to prove that for any rational number p/q, f(p/q)=6*p/q-5 and this allows you to plug in the value -17/6 and you get the result. You have to prove that fir a postive number n, f(n*x)=n*f(x)+(n-1)*5. From this, you then you use x=1/n and deduce the forumala for f(1/n), then f(p/q) for positive rational numbers and then you prove that the forumal also holds for negative rational numbers by calculating calculating f(0)=-5 nd then calculating f(x-x) to express f(-x) in terms of f(x)
Note that for irrational numbers, the initial conditions are not sufficiant to completely define the function.
if you have f(1/n) then easy to compute f(-3)+f(1/6)
Assume f(x) to be linear f(x)=ax+b. Substituting in original equation
a(x+y)+b=ax+b+ay+b+5 So b = -5
f(x)=ax-5 Substitute x=1 gives f(1)=a-5=1 so a=6
f(x)=6x-5 and verifies assumption of linearity.
Put x=-17/6 then f(-17/6)=-22
Nice!
f(x) appears to be linear. From f(x+y)=f(x)+f(y)+5 and f(1)=1, one gets immediately f(x) = 6x - 5.
There are actually infinitely many nonlinear solutions, too. It is true that f(x)=6x-5 for rational x, though.
Show 👏your 👏work👏
@@MathFromAlphaToOmega You are incorrect . Or rather should I say, using terminology wrongly and using ill-defined terms. Frankly,😄and this will seem insanely absurd... None of the solutions is in fact "Linear"!!! 🤯
Or perhaps, according to some other reply of yours, you are unintentionally referring to the wrong example, but still, you are using an ill-defined term.
@@UA-cam_username_not_found Yes, they are linear. Why do you say that they aren't?
@@MathFromAlphaToOmega I say that because, they simply😐aren't.
Let K be a field and V and W be K-vector spaces
A function f: V→W is linear if and only if it satisfies the properties f(v1+v2)=f(v1)+f(v2) for all v1, v2 in V and f(kv)=kf(v) for all v in V and k in K , ie a function that preserve the vector space structure. in particualar V , W and K can be equal to R.
This means the functions f : R→R such that f(x) = ax + b are in fact nonlinear as they fail one of the properties (both of them in fact)
Instead, those are called "affine functions" , a composition of a linear function and a translation. That's what I meant by "incorrect terminology".
Now, according to another comment of yours, I thought you are referring to the solutions to Cauchy's equation, which we obtain from the equation in the video after doing a change of the unknown function, g = f + 5 (here, 5 stands for the constant function that gives 5 everywhere). In that case you would be correct to say that there are linear solutions and nonlinear solutions .... Eeexcept that the term "linear" is not well-defined on its own.
Going back to the defintion I gave above, K can be any field, if we choose V and W to be R, K can be in particular Q or R, and we can see that the set of linear function R→R will differ if we change K into different fields, and thus we should also specify this detail and call the functions K-linear.
We can check that all the solutions to Cauchy's equation where the unknown is a function R→R are in fact Q-Linear, but only the monotonic solutions are R-linear.
Quicker with the remark that -17/6 = -3 + 1/6. f(-3) is deduced from f(3), itself deduced from f(2) and f(1).
f(1/6) is deduced from f(1/3) which is deduced from 1/3 + 2/3.
Great video. Greetings from Colombia.
define the following function:
g(x) = f(x) + 5
Then g(x+y)=f(x+y) + 5 = f(x) + 5 + f(y) + 5 = g(x) + g(y)
that means g is linear so we know that: g(x) = ax
In particular g(0) = 0.
now go backwards: f(x) = g(x) - 5 = ax - 5
in particular: f(0) = -5
and using the fact that f(1) = 1: a - 5 = 1 so a = 6
so f(x) = 6x - 5
Wow! Nice
The equation g(x+y)=g(x)+g(y) does not necessarily mean that g is linear. Look up "Cauchy's functional equation" for some counterexamples.
@@MathFromAlphaToOmega I guess I assumed that f (and then g also) is continuous, in that case all solutions are linear.
But you know that's not a necessary assumption is we restrict ourselves to Q.
Nothing in the formulation of the problem says we have to work on R.
And over Q all solutions are linear.
Pongo f(x)=ax+b lineare...dai dati con x=0,y=1 risulta f(1)=f(0)+f(1)+5...f(0)=-5...a questo punto con f(0)=-5 e f(1)=1,trovo a e b....risulta f(x)=6x-5...f(-17/6)=-22
We're told that f(1)=1 and f(x+y)=f(x)+f(y)+5. Let y=1; then we get that f(x+1)=f(x)+f(1)+5, or f(x+1)=f(x)+6. Clearly f(x) is linear. Also, f(2)=f(1)+f(1)+5=1+1+5=7. The function is of a straight line passing through (1,1) and (2,7). Using the formula for the equation of a line passing through two points, y-1=((7-1)/(2-1))*(x-1), or y=6x-5.
There are actually infinitely many nonlinear solutions, too. It is true that f(x)=6x-5 for rational x, though.
Set y = nx. Then f((n+1)x) = f(x) + f(nx) + 5
Next, we'll take a few values.
f(x) = 1f(x) + 0
f(2x) = 2f(x) + 5
f(3x) = 3f(x) + 10
Pattern seems to be that f(nx) = nf(x) + 5(n-1). Let's apply induction to prove this. Base case is already covered above.
Assume true for n=k. Therefore we know that f(kx) = kf(x) + 5k - 5. Consider n=k+1. That is, we want to prove that f((k+1)x) = (k+1)f(x) + 5k. But we already know, from the recursion formula above, that f((k+1)x) = f(x) + f(kx) + 5. Substitute in f(kx) = kf(x) + 5k - 5, and we get f((k+1)x) = f(x) + kf(x) + 5k - 5 + 5 = (k+1)f(x) + 5k. Which is what we wanted to prove, so we're good.
Now, we are given that f(1) = 1. From our formula, we know that f(6/6) = 6f(1/6) + 5*6 - 5. So, 1 = 6f(1/6) + 25. So, 6f(1/6) = -24, or f(1/6) = -4.
Next, we know that f(17/6) = 17f(1/6) + 5*(17-5) = 17(-4) + 5*16 = -68 + 80 = 12.
Finally, we need to figure out the negative. First, note that if x = y = 0, we get f(0) = f(0) + f(0) + 5, or f(0) = -5. Next, if we set y = -x, we get f(0) = f(x) + f(-x) + 5. Since f(0) = -5, we get -5 = f(x) + f(-x) + 5, or f(-x) = -10 - f(x). So, f(-17/6) = -10 - 12 or -22.
Hence, solution is f(-17/6) = -22.
We still don't know if f can actually exist. Are the two constraints consistent? Is f well defined? A closed form for f would answer this question. What you have shown is that, if there is such a function f, this must be the value of f(-17/6). But I'm not convinced f exists.
Try f(x) = 6x-5
Is this comment as dumb as i think it is or do i not get something lmao... The solution we found (6x-5) works so yes obviously f exists.
@@ilias-4252why are you so pissed off because of a comment in a math video? He just didn't understand
@@ilias-4252nowhere in the video I see f explicitly calculated.
Or just assume a linear function: f(x) = ax + b. You will get a and b from the two conditions f(x + y) = f(x) + f(y) + 5 and f(1) = 1.
f(x + y) = f(x) + f(y) + 5
knowing that x and y are independent, you can take the derivative of both sides with respect to x
f'(x + y)derivative(x + y) = f'(x)(derivative of x) + f'(y)(derivative of y) + 0
also the derivative of x (with respect to x) is 1
the derivative of y is zero
the derivative of x + y is 1
so you get:
f'(x+y) = f'(x)
no matter what the value is of x or y, this is always true
the only way this can be true is if f'(x) = some_constant
let's call this constant "m"
f'(x) = m
integrate both sides with respect to x to get:
f(x) = mx + b
in other words, it has to be a linear function. Once you know this must be true, the equation becomes very easy to solve.
There are other solutions that aren't differentiable, though.
I solved it differently. From f(x+y)=f(x)+f(y)+5, i supposed f is linear. So I tried to find f(x) as ax+b. From the expression of f(x+y), I could equate a(x+y)+b to ax+b+ay+b+5, which leads to b=-5. Using now the condition f(1)=1, I found a=6. Hence f(x)=6x-5, and you get f(-17/6)=-22 by substitution
f(x)=6x-5
f(-17/6) // Question
f(1)=1 // A premise
f(x+y)=f(x)+f(y)+5 // B premise
f(1+0)=f(1)+f(0)+5 // C setting up to solve for f(0) using B
f(1)=1=1+f(0)+5 // D plugging A into C
1=6+f(0) // E simplifying D
f(0)=-5 // F solving for f(0) using E
f(1+1)=f(1)+f(1)+5 // G Setting up to solve for f(2) using B
f(2)=1+1+5 // H Plugging in A into G
f(2)=7 // I Simplifying H
f(2+1)=f(2)+f(1)+5 // J Setting up to solve for f(3) using B
f(3)=7+1+5 // K Plugging in A and I into J
f(3)=13 // L Simplifying K
f(2+2)=f(2)+f(2)+5 or f(1+3)=f(1)+f(3)+5 //M Setting up to solve for f(4) using B via two different methods
f(4)=7+7+5 or f(4)=1+13+5 //N Plugging in I into M to solve for method 1 and Plugging in A and L into M to solve for method 2
f(4)=19 // O Simplifying both methods in N resulting in same result
{(0,-5),(1,1),(2,7),(3,13),(4,19)} // P Establishing A, F, I, L, and O as ordered pairs, under new observation that this system is linear
y=mx+b // Q known equation: Slope-Intercept form of a line
m=(1-(-5))/(1-0)=6 // R Establishing slope from two points in P
y=6x-5 // S Plugging slope from R and y-intercept point from P into Q
f(x)=6x-5 // T Establishing S as a function
f(-17/6)=6(-17/6)-5 // U Plugging in -17/6 into T
f(-17/6)=-17-5 // V Simplifying U
f(-17/6)=-22 // W Solved
*slight edit* forgot to add in I on step P
f(x+y)-f(x)=f(y)+5, now at y=0, f(0)=-5 =>lim y->0 (f(x+y)-f(x))/y=lim y->0 (f(y)+5)/y... now LHS= f'(x), RHS=f'(0)=constant,, since f(x) is a function hence only 1 value of f(x) at any 'x', hence only one value of f'(x) at that 'x'. hence f'(x)=k(const) => f(x)=kx+c... now,, putting [f(x),x]≡[1,1], [0,5]=> c=-5, k=6 => f(x)=6x-5.
That's assuming f is differentiable, which it may not be.
@@MathFromAlphaToOmega f'(x)=f'(0) which means at every point the slope is same f'(0).. now if f'(0) is undefined then it means f(x) is not differentiable at every point in domain.. which is impossible for a continuous function.
@@mrityunjaykumar4202 There are continuous functions that aren't differentiable anywhere (the Weierstrass function). Also, the nonlinear solutions to f(x+y)=f(x)+f(y)-5 aren't continuous anyway.
@@MathFromAlphaToOmega The Weierstrass function is not monotonically inc or dec...this means there are integers such that value of function at those integers is lesser than at integers lesser in magnitude ..but in the functional equation f(x+y)=f(x)+f(y)+5, at x=x and y=1.. f(x+1)=f(x)+1+5.. this shows monotonically increasing for integers.. hence its not Weierstrass function and is differentiable at some points in the domain.
@@mrityunjaykumar4202 There are many functions that aren't differentiable anywhere. The Weierstrass function is just one of them. The other solutions to f(x+y)=f(x)+f(y)-5 aren't continuous anywhere, let alone differentiable.
odvajamo po x: f'(x+y) = f'(x)
odvajamo še po y: f''(x+y) = 0 in vstavimo y = 0: f''(x) = 0
f'(x) = a
f(x) = ax + b
f(x) = f(x) + f(0) + 5 : f(0) = -5 = b in f(1) = a - 5 = 1: a = 6
f(x) = 6x-5
f(-17/6) = -17-5 = -22