Jordan's Inequality Visual Proof

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This is a short, animated visual proof demonstrating Jordan's inequality, which is an equality providing upper and lower bounds for the sine function over the interval [0, pi/2].
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For other inequality visual proofs, see my inequality playlist:
• Inequalities
This animation is based on a visual proof by Feng Yuefeng from the April 1996 issue of Mathematics Magazine (www.jstor.org/... page 126 ).
#math #inequality #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #algebra #areas #mathematics #jordaninequality #jordan #algebraicidentity #mathshorts #mathvideo #mtbos #circle #unitcircle
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Опубликовано:

14 окт 2024

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Комментарии : 40

@wirebrushproductions1001 4 месяца назад

The inequality also reduces to "pi is equal to or greater than 2".

@samval5 5 месяцев назад

Look at the sin curve. The inequality results of the position of the curve between the cord between 0 and pi/2 and the tangent line at 0

@alanthayer8797 5 месяцев назад

Visual visuals VISUALS!

@MathVisualProofs 5 месяцев назад

👍😀

@Jamesbondhere 5 месяцев назад

Wonderful proof!! Although u said it's not ur proof, u explained it well!!

@MathVisualProofs 5 месяцев назад

Thanks!

@OtherworldlyYTP 5 месяцев назад

Thanks for making this!

@MathVisualProofs 5 месяцев назад

Thanks for checking it out!

@nicopb4240 5 месяцев назад

Truly amazing, thanks

@MathVisualProofs 5 месяцев назад

Thanks for watching!

@STEAMerBear 5 месяцев назад

What are some useful applications or implications of Jordan’s inequality?

@Aditya_196 5 месяцев назад

Well a question would be great... Prove : sin(sin(x))*x ≥ sin²(x) for X belongs to 0 to π/2

@STEAMerBear 5 месяцев назад

@@Aditya_196 THANK YOU!!!

@Aditya_196 5 месяцев назад

@@STEAMerBear solved it ?!

@gameplay-masterstudios4305 5 месяцев назад

I liked this vedio so that RU-vid will recommend this vedio to someone like me

@MathVisualProofs 5 месяцев назад

Thanks!

@mekbebtamrat817 5 месяцев назад

I think the inequality was 'slipped in there' when the inequality about OB and OA was mentioned. We do not exactly prove that and we simply take it as a given that OB falls outside of the unit circle. We would also not know if the diametrically opposite point to B lies behind O or not.

@philipyao5989 5 месяцев назад

since the inner circle intersects the unit circle at P and Q, and circles can intersect in at most 2 points, that means the part of the inner circle to the right of the unit circle must be entirely outside of the unit circle, and similarly with the left side.

@mekbebtamrat817 5 месяцев назад

@@philipyao5989 Ah yes, that is a great explanation.

@joshmguni 12 дней назад

за такие доходные связки скоро бинанс за тобой выедет))))

@turtlgamez 5 месяцев назад

I can’t wait until I’m in classes that let me understand these

@grpthry4659 5 месяцев назад

How did you get that MP is always greater than MA? This doesn't seem obvious to me. Also, similar question on showing that arclength of PB being greater than that of PA.

@MathVisualProofs 5 месяцев назад

It’s really just that OB is greater than OA. Since the arcs connect P and Q we just need B father away from origin. This also only works for angles in first quadrant.

@maxchemtov3482 5 месяцев назад

Draw a line segment L connecting A to the top of the original unit circle. Notice that L is completely contained inside the unit circle (since it's convex). Let R be the intersection of L with MP. Then MR = MA (since MRA is a 90-45-45 isoceles triangle) and MP = MR + RP (since R is inside the circle). Putting these together, MP = MA + RP >= MA.

@_P_a_o_l_o_ 5 месяцев назад

How does the inequality between the arc lengths follow from the previous inequality?

@MathVisualProofs 5 месяцев назад

Both arcs connect P and Q so we want to make sure that B is farther from the origin than A.

@typha 5 месяцев назад

1:13 Why is it that you're concluding PBQ >= PAQ?

@MathVisualProofs 5 месяцев назад

Both arcs from P to Q and B is farther away from origin than A.

@typha 5 месяцев назад

@@MathVisualProofs That a path is further away from the origin than another path does not mean that it is the longer one.

@blblblblblbl7505 2 месяца назад

@@typha I think for convex curves like in this case the argument should be valid. Specifically, if two convex curves have the same endpoints and don't cross, then the curve that is further from the straight line connecting those points is the longer curve. You can prove it for polygonal paths first, and then the general case can be proved by approximating the curves with polygonal paths.

@typha 2 месяца назад

@@blblblblblbl7505 perhaps such a lemma would have been useful, and it's proof would have likely been more complicated than the rest of the two and a half minute video.

@blblblblblbl7505 2 месяца назад

@@typha Agreed. I think the idea is it should be "visually obvious" that the outside one is bigger. That said, seems like you could just draw the graph of y = sinx and the two lines and this will also prove it, if the idea is just to give a visually simple proof.