@@neevhingrajia3822 Let me rephrase myself. We're not actually substituting 𝑡 = 𝑥, we are just replacing 𝑡 with 𝑥 as the input variable of 𝑓. The reason we can do this is that the range of 𝑥 (i.e., the set of all positive real numbers) is equal to the range of 𝑡 = 1 ∕ 𝑥, so replacing 𝑡 with 𝑥 doesn't change the domain of 𝑓, it is still defined for all positive real numbers.
i didn´t understand the final swich back T, instead of just replacing all T for X, shoudn´t replace it for (1/X)? wich would give us f(t)=x+x[1+(1/x)^2]^(1/2)
Think it like this way @@davidbechor3776 u got the f(t) = 1/t(1+√1+t^2) if u replace t with any real number then u will get a output. now replace the t with x now if u replace the x with any real numbers then u will get the exact same value so here "t" is only a letter it doesn;t matter that much i might make any mistake so please consider the mistake and please provide right answer.
@@davidbechor3776he used t to see what the function does to it. What the function does to z or y or x or t is the same, saying t=x+1 in the beginning doesn't apply to the end. That's why he can replace t with x at the end
What if we take derivative of both sides to dx…would it not simplify it? And then final find f of x by integrating the right side of equation…would this not work?
Good morning sir. I have a little ambiguity about the final answer. In the steps preceding, shouldn't there be a direct square root for both t^2s and the square root of 1also 1?
That is the first thing I thought. It should just be the reciprocal subbed into the original equation but in his answer he ends up with an extra factor on the left term.
@@mrnogot4251 he simplifies the radical and takes out 1/√t² as a factor then introduces an absolute value 1/|t|. But what if i dont take out a factor, will it be wrong?
We have 1 ∕ 𝑥 + 1 ∕ 𝑥⋅√(1 + 𝑥²). Multiplication takes precedence over addition, so we can't add 1 ∕ 𝑥 + 1 ∕ 𝑥 = 2 ∕ 𝑥 and then multiply by √(1 + 𝑥²). Instead, what we can do is to factor out 1 ∕ 𝑥, which gives us 1 ∕ 𝑥⋅(1 + √(1 + 𝑥²)). And since multiplying by 1 ∕ 𝑥 is the same as dividing by 𝑥, we can then rewrite that as (1 + √(1 + 𝑥²)) ∕ 𝑥.
if i am not wrong. the x at the end is when x is used to describe a function. the x at the start is an unknown. yes it is confusing, but as far as i know you should just think of the x at the start as just any other variable.@@jairogen90
@@znhaitIn functional notation, f is the name of the function and x the input space of the function. As a whole f(x) is the output an explicit relation in x.A function does not depend on the name f,x,f(x) but on the relation between x and f(x). It is possible to say things like y=y(x) which confuses people but it just means that y depends only on x. It is weird because y is both a variable and the name of a function. I personally think that it is bad nasty notation but it is used a lot so…
I don't understand why the new variable t is even needed. Why shouldn't it be possible to stay with x and just introduce the inverse into the function?
If x is allowed to be negative, the expression derived in this video for f(x) won't work. It's given f(1/x)=x+√(1+x²) -- (i) Suppose x is allowed to be negative. let x=-1. (i) gives: f(1/-1)=-1+√[1+(-1)²]=-1+√(1+1)=-1+√2 i.e, f(-1)=-1+√2 -- (ii) In the video, he derives f(x)=(1/x)[1+√(1+x²)] -- (iii) This can be used to evaluate f(-1) Putting x=-1 in (iii), we get: f(-1)=(1/-1)(1+√[1+(-1)²])=-(1+√2)=-1-√2 -- (iv) You can notice that RHSs of (ii) and (iv) are different
You are right. The only problem that arises is when we try to simplify the expression we got for 𝑓(𝑥). The best we can do is basically 𝑓(𝑥) = 1 ∕ 𝑥 + 1 ∕ |𝑥|⋅√(1 + 𝑥²), unless we want to write it as a piecewise function: 𝑥 < 0 ⇒ 𝑓(𝑥) = 1 ∕ 𝑥 + 1 ∕ (−𝑥)⋅√(1 + 𝑥²) = (1 − √(1 + 𝑥²)) ∕ 𝑥 𝑥 > 0 ⇒ 𝑓(𝑥) = 1 ∕ 𝑥 + 1 ∕ 𝑥⋅√(1 + 𝑥²) = (1 + √(1 + 𝑥²)) ∕ 𝑥
√(𝑡²) is the _positive_ square root of 𝑡², so usually we can't say √(𝑡²) = 𝑡 because if 𝑡 < 0 then 𝑡 is the _negative_ square root of 𝑡². The easy way to get around this is to say √(𝑡²) = |𝑡|, because this way the equation holds for negative values of 𝑡 as well. However, in our case the domain is 𝑡 > 0, which means that 𝑡 is not allowed to be negative, and therefore we can write √(𝑡²) = 𝑡.
Anything divided by zero has "infinitely many solutions" in fact you can choose any number to be your answer however a number sequence or function that goes to infinity got many solutions but only one answer
Never stop learning. Those who stop learning stop living.❤❤
You’re lucky this ain’t Insta
Great lines for life .
@@ramizhossain9082you are right my brother
@@a0fefdhow is that relevant to the comment in any way?
Thank you for your excellent instruction. This stuff now almost seems easy.
Nice and easy
Thank you for you awesome contributions. I am sensing an error at video position 5:10. Shouldn't the substitution of 1/t = x and not 1/x?
At this point we're substituting 𝑥 = 𝑡, not 𝑥 = 1 ∕ 𝑡
ㄷ@@jumpman8282
@@jumpman8282huh? But does that not contradict our previous assumption?
@@neevhingrajia3822
Let me rephrase myself. We're not actually substituting 𝑡 = 𝑥, we are just replacing 𝑡 with 𝑥 as the input variable of 𝑓.
The reason we can do this is that the range of 𝑥 (i.e., the set of all positive real numbers) is equal to the range of 𝑡 = 1 ∕ 𝑥,
so replacing 𝑡 with 𝑥 doesn't change the domain of 𝑓, it is still defined for all positive real numbers.
Fun tact
These two substitions can be used to remove radical sqrt(x^2+1) from integral
R(x,sqrt(x^2+1))
I need to know more
Very well done . Loved it .
In my teacher's teaching, She never gave questions as great as this😀
Do keep 'em coming!
😊
Is one over t not supposed to be x?
Bom trabalho, cara! Você tem o dom de ensinar!
i didn´t understand the final swich back T, instead of just replacing all T for X, shoudn´t replace it for (1/X)? wich would give us f(t)=x+x[1+(1/x)^2]^(1/2)
Me neither... I think he made a HUGE mistake, and nobody seems to correct him. Indeed, if T=1/x, how can you replace T with X?
Think it like this way
@@davidbechor3776 u got the f(t) = 1/t(1+√1+t^2)
if u replace t with any real number then u will get a output.
now replace the t with x
now if u replace the x with any real numbers then u will get the exact same value
so here "t" is only a letter it doesn;t matter that much
i might make any mistake so please consider the mistake and please provide right answer.
@@davidbechor3776 You can always put the same number into both equation and see if the values come out to be the same.
@@davidbechor3776he used t to see what the function does to it. What the function does to z or y or x or t is the same, saying t=x+1 in the beginning doesn't apply to the end. That's why he can replace t with x at the end
|t| = t since x > 0. So, you can remove the absolute value sign.
good chalk/board/writing................
Prime Newtons is awesome! 😊
Блин, прекрасный учитель и прекрасный канал! Рад, что случайно нашел.👍
Very good explanation.
You explain nicely.
It's more simple : f(x)= f(1/(1/x)) = (1/x)+(1/x)sqrt(1+X^2) no operations to do 😁
Very good 👍
nice video, just one question, shouldn't f(x) = f(1/t)?
What if we take derivative of both sides to dx…would it not simplify it? And then final find f of x by integrating the right side of equation…would this not work?
Hi newton, please make more videos on 'Tetration~Exponentiation Equations', as most of us like it!😊
Good morning sir. I have a little ambiguity about the final answer. In the steps preceding, shouldn't there be a direct square root for both t^2s and the square root of 1also 1?
To avoid confusion I will say, find f(y) (y > 0) where y = 1/x or x = 1/y.
Thus, f(y) = (1/y) + √(1 + 1/y^2) = [1 + √(y^2 + 1)]/y
Now, change y to x, giving f(x) = [1 + √(x^2 + 1)]/x
👍👍👍👍👍👍👍
We make sure that t > 0 or take the absolute value
Sorry I found out you're absolutely correct. Thanks
Why can't I leave the answer as f(x) = 1/x +√(1 + 1/x²) ?
That is the first thing I thought. It should just be the reciprocal subbed into the original equation but in his answer he ends up with an extra factor on the left term.
@@mrnogot4251 he simplifies the radical and takes out 1/√t² as a factor then introduces an absolute value 1/|t|. But what if i dont take out a factor, will it be wrong?
It's perfectly fine to leave it like that. This way the function also holds for 𝑥 < 0.
Why isn't it 1/x(2+sqr. Root of 1+x^2.. Since 1/x+1/x is = 2/x
Plsss I need answer..
Correct me if I'm wrong
We have 1 ∕ 𝑥 + 1 ∕ 𝑥⋅√(1 + 𝑥²).
Multiplication takes precedence over addition, so we can't add 1 ∕ 𝑥 + 1 ∕ 𝑥 = 2 ∕ 𝑥 and then multiply by √(1 + 𝑥²).
Instead, what we can do is to factor out 1 ∕ 𝑥, which gives us 1 ∕ 𝑥⋅(1 + √(1 + 𝑥²)).
And since multiplying by 1 ∕ 𝑥 is the same as dividing by 𝑥, we can then rewrite that as (1 + √(1 + 𝑥²)) ∕ 𝑥.
I like your presentation style. Are you sure your substution back to x is correct?
It is not a substitution. The last x is not the original x.
@@PrimeNewtons that's confusing
if i am not wrong. the x at the end is when x is used to describe a function. the x at the start is an unknown. yes it is confusing, but as far as i know you should just think of the x at the start as just any other variable.@@jairogen90
@@PrimeNewtons I am decent at math, and it can be hard to convince myself of dummy variables. I think that's the question that is being asked.
@@znhaitIn functional notation, f is the name of the function and x the input space of the function. As a whole f(x) is the output an explicit relation in x.A function does not depend on the name f,x,f(x) but on the relation between x and f(x). It is possible to say things like y=y(x) which confuses people but it just means that y depends only on x. It is weird because y is both a variable and the name of a function. I personally think that it is bad nasty notation but it is used a lot so…
Teacher, you made a mistake in the substitution of 1/x in square t into the root
❤
I don't understand why the new variable t is even needed. Why shouldn't it be possible to stay with x and just introduce the inverse into the function?
Not impossible. Just a different strategy
@@PrimeNewtons Thanks for the answer. But what does it help to switch to t anyway? Sorry, but I am not a mathematician.
Can you do this with Q
For the domain surely you can have x > 0 or x < 0 since the x in the square root is squared so the negative doesn’t cause a problem
If x is allowed to be negative, the expression derived in this video for f(x) won't work.
It's given f(1/x)=x+√(1+x²) -- (i)
Suppose x is allowed to be negative. let x=-1. (i) gives:
f(1/-1)=-1+√[1+(-1)²]=-1+√(1+1)=-1+√2
i.e, f(-1)=-1+√2 -- (ii)
In the video, he derives f(x)=(1/x)[1+√(1+x²)] -- (iii)
This can be used to evaluate f(-1)
Putting x=-1 in (iii), we get:
f(-1)=(1/-1)(1+√[1+(-1)²])=-(1+√2)=-1-√2 -- (iv)
You can notice that RHSs of (ii) and (iv) are different
That’s because the function might not have an inverse which is problematic.
You are right. The only problem that arises is when we try to simplify the expression we got for 𝑓(𝑥).
The best we can do is basically 𝑓(𝑥) = 1 ∕ 𝑥 + 1 ∕ |𝑥|⋅√(1 + 𝑥²),
unless we want to write it as a piecewise function:
𝑥 < 0 ⇒ 𝑓(𝑥) = 1 ∕ 𝑥 + 1 ∕ (−𝑥)⋅√(1 + 𝑥²) = (1 − √(1 + 𝑥²)) ∕ 𝑥
𝑥 > 0 ⇒ 𝑓(𝑥) = 1 ∕ 𝑥 + 1 ∕ 𝑥⋅√(1 + 𝑥²) = (1 + √(1 + 𝑥²)) ∕ 𝑥
Hey, guy you are damn smart.
Last step is incorrect. You had to switch t To be (1/x).
Because 1/x is an involution, f(x) is simply f(1/1/x)
🦹♂🎈االله
Cant we just give x the value of 1/x and find f(x)??
t=1/x
sir.. f(X) =DEFFERENT ANSWER
Agree
Great, Handsome.
🇮🇹
It must be (1/x)^2
i didnt understand the |t| part. can u explain again
√(𝑡²) is the _positive_ square root of 𝑡², so usually we can't say √(𝑡²) = 𝑡 because if 𝑡 < 0 then 𝑡 is the _negative_ square root of 𝑡².
The easy way to get around this is to say √(𝑡²) = |𝑡|, because this way the equation holds for negative values of 𝑡 as well.
However, in our case the domain is 𝑡 > 0, which means that 𝑡 is not allowed to be negative, and therefore we can write √(𝑡²) = 𝑡.
@@jumpman8282 okay thank you I understand it now
The absolute value sign accounts for keeping both negative and positive values that t can be
I think it is wrong....t = 1/x so x = 1/t.....
😅no it not wrong
x must be greater than zero because if you put x=0 the function becomes infinite. Any thing divided by zero is infinity.
@IonRuby please correct me brother. I am sure that this is the reason why x not be 0. Similar case apply for all rational numbers.
@@SalmanKhan-qp5gzI assume he’s objecting that 1/0 is undefined, not infinity.
It is undefined. You have to use a limit sign for it to be infinity or negative infinity.
limit does not exist for 1/x as x -> 0 because 0+ and 0- are different@@KingGisInDaHouse
Anything divided by zero has "infinitely many solutions" in fact you can choose any number to be your answer however a number sequence or function that goes to infinity got many solutions but only one answer
En the end one misteake .
f(x)=((1+(1+x^0,5))/x