# Graph sketching

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The A.G

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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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mqb2766

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#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! + ...

Not sure about the x->0 part of your sketch if its

(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.

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! + ...

Not sure about the x->0 part of your sketch if its

(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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The A.G

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#3

(Original post by

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! + ...

Not sure about the x->0 part of your sketch if its

(x + sin(x))/(x-sin(x))

It will be something like ~ 1/x^2

**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! + ...

Not sure about the x->0 part of your sketch if its

(x + sin(x))/(x-sin(x))

It will be something like ~ 1/x^2

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mqb2766

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#4

(Original post by

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 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?

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.

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DFranklin

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#5

(Original post by

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! + ...

**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! + ...

In particular, why shouldn't you consider x - x^3/3! + x^5/5! -... which clearly is bigger than x for large x?

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mqb2766

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#6

(Original post by

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?

**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?

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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#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.

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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DFranklin

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#8

(Original post by

The expansion was only for small x (obviously).

**mqb2766**)The expansion was only for small x (obviously).

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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(Original post by

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...

**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...

Last edited by mqb2766; 1 month ago

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#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,...

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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