why sin(x)≈x, (i.e. the famous small-angle approximation)
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- Опубліковано 24 бер 2022
- This is perhaps the most famous linearization of a function. Yes, the small-angle approximation for sin(x). This is a very popular approximation in physics and engineering so you should definitely know how we get this result!
Related videos:
This is how to get the exact value for sin(pi/18) • this special triangle ...
How small is "small" when we use the small-angle approximation? • The Small Angle Approx...
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Related videos:
This is how to get the exact value for sin(pi/18) ua-cam.com/video/_00oskWLtII/v-deo.html
How small is "small" when we use the small-angle approximation? ua-cam.com/video/ppN7pVQaFhY/v-deo.html
The fundamental theorem of engineering! :D
I thought that "assume it's round" was the main physics assumption
Ah yes. π = 3 and e = 3 and π^2 = g = 10.
@@ausaramun yes, it's facts
Wrong. The fundamental theorem of engineering is "sin(x)=x" not "sin(x)≈x"
@@monika.alt197 well in engineering =≈=
So yeah
I feel like this video is gonna just have 70% of the comments being about the fundamental theorem of engineering.
This feels like a crime in math but physics says it's legal and satisfying 😂
Excellent video, thanks! This is great and used in physics all the time.
This also works for tan(x)*, because tan(x) is just sin(x)/cos(x), and at 0, cos(x)=1 and 1/cos²(x)=1. Or just at 0, d/dx sin(x) = d/dx tan(x) = d/dx x = 1. So, it works.*
*For SMALL values of x (even smaller than for sin(x)). Like, say, x=1 arcsecond.😝
Ah, the fundamental theorem of engineering!
Since we know that sin(x) is approximately equal to x and that it gets more and more accurate for smaller angles x:
sin(x)~=x
-> sin(pi/x)~=pi/x
sin(pi/x)*x~=x
So when we take the limit for x approaching infinity, we get that sin(pi/x)*x=pi
Pretty cool. Please,is it possible to evaluate sinx using 'Weierstrass Substitution?
can you do the same for tan α ≈ α and cos α ≈ 1 ? thank you
Adding the second term of the Taylor series makes it even more accurate, an even more accurate estimate (albeit more complicated, perhaps too complex for an engineer) is x- (1/6)x^3
or with taylor series :
sin(x) = x-x^3/3!+x^5/5! - x^7/7!+ ... , with small x, i can expand taylor series with 1 term because with small angles O(x^3) is so small (u can expand more ... or use a calculator aproximation ... )
sin(x) ≈ x
Bonus :
cos(x) = 1-x^2/2! + x^4/4! -x^6/6! + ... , with small x, we (physicist) usually expand for 2 terms (...this is what my classical mech professor said) , cos(x) ≈ 1-x^2/2! (for small x)
3:50 I thought π in this case is in radian ie 180°
hmm what about this : at what value of x does sin(x) ≈/≈ x anymore
The biggest problem of this proof is that you need to differentiate sine function first. But to find the derivative of sine function, you need lim x->0 (sinx/x )= 1. from the first principle.
Therefore I do not agree with your argument.
Edit: I changed back to 1. My mistake.
????
he did at 1:36, so QED
@@bprpcalculusbasics It makes sense to me. The derivative of sin x requires that (sin x)/x = 1 as x approaches 0. This means that sin x ~ x for small enough x.
So it seems like the linearization of sin x at x=0 is required to determine the derivative of sin x in the first place.
@@ZipplyZane u can do a geometric proof for that limit.
"lim x->0 (sinx/x )= 0" is nonsense. It's equal to 1
I thought this approximation was because 'x' is the first-order in the sin(x) expansion in the infinite series?
Second question....if cos(x) is approximated would it be just 1, or would it be (1- q²) / 2! ?
For this, I thought the approximationwould need the first q value, as opposed to just taking the approx as just 1. Thoughts?
cos(x) = 1 is the best _constant_ (and also the best linear) small-angle approximation for cosine. cos(x) = 1 - x^2 / 2 is the best _quadratic_ (and also the best cubic) small-angle approximation. How many terms you need to represent the approximation is up to you.
The linear approximation (linearization) takes the first two values in the function's expansion which is equivalent to what has done in the video, the quadratic approximation takes the first three terms. Since the cosine function has constant two first terms then the approximation will be as quadratic approximation.
I think you can actually approximate it like that until x = 30°
Going to have an important calc 1 test soon, does anybody have useful tips?
natural logarithmic and exponential functions are really important for calc 1, I advise you to focus on those
@@kepler4192 Ok, thanks
@@nadva304 np
@@kepler4192 ....I thought those were the first week in calc 2. Granted everyone's calc classes will likely vary.
@@FrogworfKnight I’m in highschool and I learnt those last semester
Hello!
taylor series!
Yeah this is great, sin x = x, sinh x = x, cos x = 1, e^x = 1!!:D
How to find the integral of x³tan²(x²) ?
x^2 = u
xdx = du/2
so ur integral becomes
u.tan^2(u)/2
Apply by parts now
u.integral(tan^2u) - integral(tan^2u) [differentiation of u is 1]
now u can use tan^2u = sec^2u-1
and integral(sec^2u) = tan(u)
Wolphramalpha
Please include an english subs next time, Im begging youuu I'm dying to learn thisss
ok
oops! He exposed us! 🤭
Are all calculus teachers magicians of math?
Magicians of calculation. Not really of math, rigorous proofs and all that stuff.
Mathematicians are
hi
x=0
sinx=0=x
*E Z*
Lmao even by engineering standards this doesnt make sense.... x stays at least two orders of magnitude larger than sinx of x at 10^-80. This approximation should only be used if youre adding or subtracting sinx for small x's. This should never be used for division or multiplication youll get orders of magnitude of error. For example 10+ 10 sinx can be 10 + 10x for small x , but 10/sinx is is never 10/x for any small x
試著用sintheta= y/r去解釋