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#1
Is it a given fact that x>sin(x) for all values of x. I was unsure so messed up my logic when evaluating at x=0. I also differentiate the curve to find the local maximum and minimum even though I knew it probably wouldn’t yield anything useful since I ended up with x=tan(x). But would it be good as I’m trying to be rigorous with my deductions
Last edited by The A.G; 1 month ago
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1 month ago
#2
x > sin(x)
for all x > 0. It can be easily seen if you think about the circle / triangle interpretation or think about the series expansion of sin(x)
x - x^3/3! + ...

(x + sin(x))/(x-sin(x))
It will be something like ~ 1/x^2 and hence a vertical asymptote at x=0.

For "large" x, it will be something like 1+sin(x)/x, so the mexican hat which is sin(x) where the magnitude goes down like 1/x.
Last edited by mqb2766; 1 month ago
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#3
(Original post by mqb2766)
x > sin(x)
for all x > 0. It can be easily seen if you think about the circle / triangle interpretation or think about the series expansion
x - x^3/3! + ...

(x + sin(x))/(x-sin(x))
It will be something like ~ 1/x^2
I see so by considering the maclaurin expansion of sin(x) about 0 I’ll be able to get the 1/x^2 since the factorial dominates the numerator so I’ll just take the first term?
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1 month ago
#4
(Original post by The A.G)
I see so by considering the maclaurin expansion of sin(x) about 0 I’ll be able to get the 1/x^2 since the factorial dominates the numerator so I’ll just take the first term?
The factorial part isnt that relevant (its a constant), its really the x^? thats important. On the top you'd have
x + x - x^3/3! + ....
and on the bottom
x - x + x^3/3! + ....
So
(2x - x^3/3! ...)/(x^3/3! ...)
(12 - x^2 ...)/(x^2 ...)
~ 12/x^2
So its heads to infinity pretty sharpish (proportional to 1/x^2) as x->0 (either positive or negative).

Note as both go to zero as x->0, you could use hopitals rule, but if you know the expansions, its basically the same.
Last edited by mqb2766; 1 month ago
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1 month ago
#5
(Original post by mqb2766)
x > sin(x)
for all x > 0. It can be easily seen if you think about the circle / triangle interpretation or think about the series expansion of sin(x)
x - x^3/3! + ...
I'd accept the expansion argument for *small* x, but not for all x>0. (I think it's *really* unobvious how sin x behaves for large x from the power series).

In particular, why shouldn't you consider x - x^3/3! + x^5/5! -... which clearly is bigger than x for large x?
Last edited by DFranklin; 1 month ago
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1 month ago
#6
(Original post by DFranklin)
I'd accept the expansion argument for *small* x, but not for all x>0. (I think it's *really* unobvious how sin x behaves for large x from the power series).

In particular, why shouldn't you consider x - x^3/3! + x^5/5! -... which clearly is bigger than x for large x?
For "large" x its 1+2sin(x)/x which is a shifted up mexican hat as per #2?
The expansion was only for small x (obviously).
In the middle x~2, you'd have to note its always positive as the denominator is never zero and you'd have to be a bit careful about where the first min/max are.
Last edited by mqb2766; 1 month ago
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1 month ago
#7
So ...
https://www.desmos.com/calculator/xyk2kkxyin

The three dashed curves
12/x^2
shows the x->0 behaviour and
1+/-2/x
are the approximately bounding envelopes for sin(x) when x is larger.

The large and small approximations blend together fairly well.
Last edited by mqb2766; 1 month ago
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1 month ago
#8
(Original post by mqb2766)
The expansion was only for small x (obviously).
Don't want to seem like I'm nitpicking,but it wasn't obvious to me from your post and even rereading it now it's still not obvious. If it was obvious to the op that's all that matters of course.

As so frequently the case I'm handicapped by being unable to read the attached image - why it's impossible to post things the right way up I don't know...
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1 month ago
#9
(Original post by DFranklin)
Don't want to seem like I'm nitpicking,but it wasn't obvious to me from your post and even rereading it now it's still not obvious. If it was obvious to the op that's all that matters of course.

As so frequently the case I'm handicapped by being unable to read the attached image - why it's impossible to post things the right way up I don't know...
Thought Id said the expansion was for x->0 and the mexican hat for large x, but it is what it is. If you mean the x>sin(x) part, then yes I "assumed" that for x large, sin was bounded by 1 rather than using the expansion argument. The expansion argument was really only meant to support the small x part. It was meant to provide quick, simple reasoning to help sketching, thats all.
Last edited by mqb2766; 1 month ago
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1 month ago
#10
The A.G As mentioned above, the main thing that was missing for me was the "middle" bit and being clear about how the stationary points of sin(x) are altered by the denominator. So for
1 + 2sin(x)/(x-sin(x))
Differentiating and setting = 0, then the stationary points will occur when
tan(x) = x
Similar to the x>sin(x) reasoning,
tan(x)>x
for x small (< pi/2) and with a bit of logic (where x intersects with tan(x)), the sin() min max locations won't be altered by that much (the new locations are slightly less), apart from the maximum at pi/2 disappears/is swamped by the 12/x^2 local approximation. The first stationary point will be a minimum at ~3pi/2 (slighltly less) and then you'll have the mexican hat function roughly bounded by 1+/-2/x with stationary points at approximately 3pi/2,5pi/2,7pi/2,...
Last edited by mqb2766; 1 month ago
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