@@kamarinelson Let me think..... cosx = 1-x^2/2+x^4/4!-x^6/!-.... cos(ix) = 1-(ix)^2/2+(ix)^4/4!-(ix)^6/6!+.... cosix =coshx ...... cos(ix) = 1 - -(x)^2/2+ +(x)^4/4! - -(x)^6/6! = 1 + x^2/2 + x^4/4! + x^6/6!... = sum(x^2n/(2n)!) I was hoping for something imaginary to pop up. Now i'm sad. :(. also the sequence makes since because coshx diverges the the ends.
i still have a confusion like if we took the other part of the curve (minus inf, 0} we will arrive at a different solution ??? @blackpenredpen plz answer this i am tempted
The horizontal line test (HLT) is used to determine whether or not the inverse of a function exists; it's the opposite of the vertical line test (VLT). HLT says that when you align a ruler with the x-axis on the graph of a function, if there are no locations where the ruler passes through multiple points at a time, the inverse exists; otherwise it does not. This comes from the fact that inverting a function is equivalent to flipping its graph over the line y = x.
@@Aruthicon to add on to your answer - we use the horizontal line test because it tests whether the function is injective (one-one). Functions that are invertible are always injective, thus the test works out (the converse; that all injective functions are invertible, is not actually true because functions need to be subjective (onto) as well)
The reason is that functions, *by definition*, must be single-valued. Otherwise, they are only relations. This restriction is in place because functions are just much more useful than relations in general, since they are completely consistent with their outputs. Relations aren't useless in themselves, though. There are probably textbook chapters dedicated to the discussion of Riemann surfaces, which go into heavy detail on the geometry of things like the square root or the logarithm over the complex plane, and particularly on the concept of "branches". The square root has two branches since the squares of negatives and positives are the same, and the logarithm has infinitely many branches because the exponential function is periodic in the imaginary direction. I personally think Riemann surfaces are very cool, but functions have their place.
The short answer is, "because we said so". By definition, function means it only has one output value, as opposed to relation which is the more general term. At least, that's the way they define a function in high school math. Some books use the term multivalued function, to refer to relations in general that may have multiple output values. Usually, function without the adjective multivalued, specifically means that there is only one output.
Can we define y values for negative x? Look: Cosh is similar to parabola. Inverse parabola is sqrt, which is defined only for x>0. We can imagine values for x
I think he meant that he'd talk about it in another video. The basic reason that cosh resembles but is not considered to be a parabola is its Taylor series: cosh(x) = 1 + 1/2 x^2 + 1/24 x^4 + .... For inputs closer to zero, the quadratic term dominates the rest of the expansion, which gives it a parabolic shape near the origin; but the higher-order terms eventually dominate as you move away from zero. Compare this to a typical quadratic equation, which has the form y = ax^2 + bx + c.
Given: cosh(i^(-1)) = m Rewrite i^(-1), by multiplying by i/i, since i don't like to be on bottom, i like to be on top. i^(-1) = -i Thus: m = cosh(-i) Rewrite cosh with its definition: cosh(x) = (e^x + e^(-x))/2 Carry out cosh(-i), which is the same as cosh(+i), by virtue of its symmetry as an even function: cosh(-i) = (e^i + e^(-i))/2 e^i = cos(1) + i*sin(1) e^(-i) = cos(1) - i*sin(1) Add them up, the i*sin(1) terms both cancel: cosh(-i) = cos(1) m = cos(1) cos(1 radian) is transcendental, and cannot be simplified further.
Given: cosh(z) = -2 Rewrite using cosh's definition: cosh(z) = (e^z + e^(-z))/2 Equate to -2, and simplify: (e^z + e^(-z))/2 = -2 e^z + e^(-z) = -4 Multiply everybody by e^z: e^(2*z) + 1 = -4*e^z Let E = e^z. Shuffle everything to the left: E^2 + 4*E + 1 = 0 Solve for E with a quadratic formula: E = -2 +/- sqrt(4 - 1) E1 = -2 + sqrt(3), E2 = -2 - sqrt(3) Both of these are negative numbers, so our solutions for z will all be complex numbers. Let z = a + b*i e^z = e^(a+b*i) = e^a * [cos(b) + i*sin(b)] In order for e^z to end up as a negative real number, this means sin(b) has to be zero, and cos(b) has to be negative. This means b can be any odd multiple of pi. cos(odd multiple of pi) = -1. Thus: e^z = -e^a, and z = a + (2*k + 1)*pi*i, where k is any integer. This means: e^a1 = -E1, and e^a2 = -E2 e^a1 = 2 - sqrt(3) e^a2 = 2 + sqrt(3) a1 = ln(1 - sqrt(3)) a2 = ln(1 + sqrt(3)) Put it together with the allowable values of b: z = ln(2 - sqrt(3)) + (2*k + 1)*pi*i z = ln(2 + sqrt(3)) + (2*k + 1)*pi*i
yes, for exactly 1 this is true, but you'll have problems finding any other real number satisfying it too. while for the '+'-version you can use any real number greater than or equal to one. besides, he decided to use the right arm, so we use the positive term. left arm, negative version. and oh boy, the place where they meet is satisfying both equations! what a miracle...
At 1, you have ambiguous case. When x=1, you are at the graphs minimum. So 1 - sqrt(x^2 - 1) >= 1 and 1 + sqrt(x^2 - 1) >= 1. For both cases, ln(1)= 0, you could not make an argument there for either case without look further out/ finding another value of x to use to eliminate the + or -. Where I would argue is the issue that ln(x) function never crosses the y- axis. For e^x, x and be any value, when x is -, y approaches 0. When x is +, y goes off to infinity. ln(x) is the inverse, so that means the minimum x value approaches zero, but cannot equal zero. So for ln(x), where x =0 or -, y does not exist. Same rule can be applied to ln(x+sqrt(x^2+1)) and ln(x-sqrt(x^2+1)). So the only case ln(x+sqrt(x^2+1))>0 is where x can be ANY value. Look as at it this way, x vs sqrt(x^2+1). No matter what, x< sqrt(x^2+1). So if x is +, then sqrt(x^2+1) will also be positive but slightly larger in magnitude. If x is -, then sqrt(x^2+1) will still be positive and larger in magnitude. That means, as long as we have x+ sqrt(x^2+1), we ALWAYS get ln(+), which will always give us a solution. (Except the case of when x= 1 as mentioned above.) Use the same logic as x-sqrt(x^2+1), you will ALWAYS get a ln(-) which does not exist.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
I like the hyperbolic functions more than the regular trigonometric functions. The way are defined is just much more satisfying!
7:32 Error 404: Speech not found
Hmmmm.... I hear curiosity killed the cat. Tayler series of cosx vs coshx. Who is mightier?!?
OK!!!!!!!!!!!!!
Correct me if I'm wrong, but isn't cos(x) just cosh(ix)? So wouldn't the forms be identical save for some alternating signs?
@@kamarinelson Let me think.....
cosx = 1-x^2/2+x^4/4!-x^6/!-....
cos(ix) = 1-(ix)^2/2+(ix)^4/4!-(ix)^6/6!+....
cosix =coshx
......
cos(ix) = 1 - -(x)^2/2+ +(x)^4/4! - -(x)^6/6!
= 1 + x^2/2 + x^4/4! + x^6/6!...
= sum(x^2n/(2n)!)
I was hoping for something imaginary to pop up. Now i'm sad. :(. also the sequence makes since because coshx diverges the the ends.
This helped me to understand hyperbolic trig functions. Thank you for the video! I have many problems to do to get this stuff memorized.
9:10 But x = 1 makes that statement true because 1 is technically greater or *equal* to 1.
yes exactly this is where I'm confused too. I was hoping he would address that here :/
You are a beautiful man and amazing at explaining
Do the sinh^-1(x) version of this :D
I already have it. You can search it on YT.
Thanks for the explanation
Very good we'll explanied dear
Wow, thanks for this. My calc book only presented the result without a proof. Now it makes more sense.
Inverse fonction of f(x)=x^3-3x-3
find the tangent line to the function y=arccosh(x), at x= 5÷4 please solve the problems
Wish you could do it using Completing the Square method ...this same one ...
Thank you for the video! It helps me a lot.
Mashallah you are so clever 🥺❤️❤️
"Calc Whatever" sounds like a very nice lecture to hear xD.
have you already done a video on how to do the taylorseries of cosh-1 or sinh-1? Nice video btw.
Not yet but I will.
Hello, How would you like to solve the cosh(z)=-1 or 0 ? ;) I suppose that the solution would be complex as real solutions do not exist... ;)
Please do a video about evaluating ζ(2) using Parseval Identity would be great ..
i still have a confusion like if we took the other part of the curve (minus inf, 0} we will arrive at a different solution ??? @blackpenredpen plz answer this i am tempted
Lifesaver, tysm
Love from india🇮🇳
Sir it also includes blue pen
So why you not include blue pen in your's channel name
Ah I forgot about the range so I was confused +ve or -ve. Thanks
so... blackpenredpen also has a blue pen
emm i am sorry , may be i have missed some of your videos , but i dont understand when you say "horizontal line test" .
The horizontal line test (HLT) is used to determine whether or not the inverse of a function exists; it's the opposite of the vertical line test (VLT).
HLT says that when you align a ruler with the x-axis on the graph of a function, if there are no locations where the ruler passes through multiple points at a time, the inverse exists; otherwise it does not.
This comes from the fact that inverting a function is equivalent to flipping its graph over the line y = x.
@@Aruthicon to add on to your answer - we use the horizontal line test because it tests whether the function is injective (one-one). Functions that are invertible are always injective, thus the test works out (the converse; that all injective functions are invertible, is not actually true because functions need to be subjective (onto) as well)
How can be the domain of cosh-¹(x) is from zero to infinity.....?
It isn't. It's from +1 to infinity. I think he's so accustomed to writing "0 to infinity" that he overlooked this one.
Can cosh^1(x)=I*theta?
What's the reason there are no functions with multiple values? I don't like that the inverse of the inverse cosh is not cosh itself.
The reason is that functions, *by definition*, must be single-valued. Otherwise, they are only relations. This restriction is in place because functions are just much more useful than relations in general, since they are completely consistent with their outputs.
Relations aren't useless in themselves, though. There are probably textbook chapters dedicated to the discussion of Riemann surfaces, which go into heavy detail on the geometry of things like the square root or the logarithm over the complex plane, and particularly on the concept of "branches". The square root has two branches since the squares of negatives and positives are the same, and the logarithm has infinitely many branches because the exponential function is periodic in the imaginary direction.
I personally think Riemann surfaces are very cool, but functions have their place.
WARNING: You'll have to go down the rabbit hole 🔬
The short answer is, "because we said so". By definition, function means it only has one output value, as opposed to relation which is the more general term. At least, that's the way they define a function in high school math.
Some books use the term multivalued function, to refer to relations in general that may have multiple output values. Usually, function without the adjective multivalued, specifically means that there is only one output.
Thanks ❤️
Can we define y values for negative x? Look: Cosh is similar to parabola. Inverse parabola is sqrt, which is defined only for x>0. We can imagine values for x
We can and we do. It turns out that arccosh(-y), where y is a positive number, is equal to arccosh(y) + i*pi
Derivative for us on your tshirt
😂
I ❤ your tshirt 😍
Yay! Thanks.
But if x is 1, you get 1...
What about the inverse hyperbolic cosecant?
I understand it finally ☺☺😍😍
#iFactOREO
you promised to discuss why cosh is not a parabola but didn't.
I think he meant that he'd talk about it in another video.
The basic reason that cosh resembles but is not considered to be a parabola is its Taylor series: cosh(x) = 1 + 1/2 x^2 + 1/24 x^4 + .... For inputs closer to zero, the quadratic term dominates the rest of the expansion, which gives it a parabolic shape near the origin; but the higher-order terms eventually dominate as you move away from zero. Compare this to a typical quadratic equation, which has the form y = ax^2 + bx + c.
Yes, that's it. Sorry, I will be teaching power series in 3 weeks. So I will have the videos then.
If x
why is e^-y times e^y equal to 1??
exp(-y)*exp(y)=exp(-y+y)=exp(0)=1 (acc. to the functional equation of the exp-function)
hi, can I get this? --------> cosh (i^-1)=m
Given:
cosh(i^(-1)) = m
Rewrite i^(-1), by multiplying by i/i, since i don't like to be on bottom, i like to be on top.
i^(-1) = -i
Thus:
m = cosh(-i)
Rewrite cosh with its definition:
cosh(x) = (e^x + e^(-x))/2
Carry out cosh(-i), which is the same as cosh(+i), by virtue of its symmetry as an even function:
cosh(-i) = (e^i + e^(-i))/2
e^i = cos(1) + i*sin(1)
e^(-i) = cos(1) - i*sin(1)
Add them up, the i*sin(1) terms both cancel:
cosh(-i) = cos(1)
m = cos(1)
cos(1 radian) is transcendental, and cannot be simplified further.
Amazing!
Gracias entendí mucho 🤣
Find a solution coshz=-2
Given:
cosh(z) = -2
Rewrite using cosh's definition:
cosh(z) = (e^z + e^(-z))/2
Equate to -2, and simplify:
(e^z + e^(-z))/2 = -2
e^z + e^(-z) = -4
Multiply everybody by e^z:
e^(2*z) + 1 = -4*e^z
Let E = e^z. Shuffle everything to the left:
E^2 + 4*E + 1 = 0
Solve for E with a quadratic formula:
E = -2 +/- sqrt(4 - 1)
E1 = -2 + sqrt(3), E2 = -2 - sqrt(3)
Both of these are negative numbers, so our solutions for z will all be complex numbers. Let z = a + b*i
e^z = e^(a+b*i) = e^a * [cos(b) + i*sin(b)]
In order for e^z to end up as a negative real number, this means sin(b) has to be zero, and cos(b) has to be negative. This means b can be any odd multiple of pi. cos(odd multiple of pi) = -1. Thus: e^z = -e^a, and z = a + (2*k + 1)*pi*i, where k is any integer.
This means:
e^a1 = -E1, and e^a2 = -E2
e^a1 = 2 - sqrt(3)
e^a2 = 2 + sqrt(3)
a1 = ln(1 - sqrt(3))
a2 = ln(1 + sqrt(3))
Put it together with the allowable values of b:
z = ln(2 - sqrt(3)) + (2*k + 1)*pi*i
z = ln(2 + sqrt(3)) + (2*k + 1)*pi*i
but 1 - sqrt(1^2 - 1) is greater than or equal to 1?
it doesn't matter, in order to prove 1 - sqrt(x^2 - 1) >= 1 as false you only need to find one counter example
yes, for exactly 1 this is true, but you'll have problems finding any other real number satisfying it too. while for the '+'-version you can use any real number greater than or equal to one.
besides, he decided to use the right arm, so we use the positive term. left arm, negative version. and oh boy, the place where they meet is satisfying both equations! what a miracle...
At 1, you have ambiguous case. When x=1, you are at the graphs minimum. So 1 - sqrt(x^2 - 1) >= 1 and 1 + sqrt(x^2 - 1) >= 1. For both cases, ln(1)= 0, you could not make an argument there for either case without look further out/ finding another value of x to use to eliminate the + or -.
Where I would argue is the issue that ln(x) function never crosses the y- axis. For e^x, x and be any value, when x is -, y approaches 0. When x is +, y goes off to infinity. ln(x) is the inverse, so that means the minimum x value approaches zero, but cannot equal zero. So for ln(x), where x =0 or -, y does not exist. Same rule can be applied to ln(x+sqrt(x^2+1)) and ln(x-sqrt(x^2+1)).
So the only case ln(x+sqrt(x^2+1))>0 is where x can be ANY value. Look as at it this way, x vs sqrt(x^2+1). No matter what, x< sqrt(x^2+1). So if x is +, then sqrt(x^2+1) will also be positive but slightly larger in magnitude. If x is -, then sqrt(x^2+1) will still be positive and larger in magnitude. That means, as long as we have x+ sqrt(x^2+1), we ALWAYS get ln(+), which will always give us a solution. (Except the case of when x= 1 as mentioned above.) Use the same logic as x-sqrt(x^2+1), you will ALWAYS get a ln(-) which does not exist.
講中文好嗎