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[FUBAR McDangles]

Could use some help on this one:

Find the x-value(s) of the relative maxima and relative minima, if any, of the function.
f(x) = x/(x+10)

So for relative extrema you find the first derivative. Through the quotient rule i figured it is 10/((x+10)^2) - which I am fairly certain is correct.

From there I am confused. For the critical values, is it just -10 because of the potential /0 scenario in the denominator? What about that 10 floating around up top?

Can anybody help me out here? I would really appreciate it.

 

[Zoned]

Hey guys, could you help me with questions like this?


Thanks.

I am assuming that you understand the basic principles of integration. To solve this question, you need to integrate cos x. Since this is an indefinite integral you will get a constant C which is being asked in the question. To get C, just put 5π/6 in the equation that you will after integrating cos x. Now let's come to how to integrate cos x to find coefficients A and B.
There is a half angle equation for cos x.

cos 2x = 2 sq(cos x) - 1.
Substitute cos x in the original question in the form of cos x/2 using the above equation. I am sure that you know how to integrate sq(cos x), use that principle.
Another identity that is useful is
sin 2x = 2cosx sinx

 

[RELAYER]

Could use some help on this one:


So for relative extrema you find the first derivative. Through the quotient rule i figured it is 10/((x+10)^2) - which I am fairly certain is correct.

From there I am confused. For the critical values, is it just -10 because of the potential /0 scenario in the denominator? What about that 10 floating around up top?

Can anybody help me out here? I would really appreciate it.

It's been a long time since I did a problem like this, but -10 would be a critical number since f'(-10) does not exist. Critical values are only when f'(x) either equals zero or does not exist, and -10 is only value for which this occurs for your derivative. So that means -10 is the only critical number. However I think -10 is neither a minima nor a maxima.

 

[FUBAR McDangles]

It's been a long time since I did a problem like this, but -10 would be a critical number since f'(-10) does not exist. Critical values are only when f'(x) either equals zero or does not exist, and -10 is only value for which this occurs for your derivative. So that means -10 is the only critical number.

See that is what I thought but I was not sure. The only example problem we had with fractions had x values in the numerator, so I thought another critical value might pop up from the numerator but rightfully not. Thanks for the help/clarification.

And I just went back into the problem and tried DNE (Does Not Exist) for relative minima maxima, just to see what would happen, and apparently there is no minima or maxima on this one. I do not understand why.

 

[Zoned]

Could use some help on this one:


So for relative extrema you find the first derivative. Through the quotient rule i figured it is 10/((x+10)^2) - which I am fairly certain is correct.

From there I am confused. For the critical values, is it just -10 because of the potential /0 scenario in the denominator? What about that 10 floating around up top?

Can anybody help me out here? I would really appreciate it.

Yes -10 will be the critical value. But this curve won't have a local maximum or minimum because if you see around x= -10, the first derivate doesn't change its sign. For x = -10 -a, FD is +ve and for x= -10+a it's still positive because of the square in denominator. Note - The a I used above is small positive value here.

 

[RELAYER]

See that is what I thought but I was not sure. The only example problem we had with fractions had x values in the numerator, so I thought another critical value might pop up from the numerator but rightfully not. Thanks for the help/clarification.

And I just went back into the problem and tried DNE (Does Not Exist) for relative minima maxima, just to see what would happen, and apparently there is no minima or maxima on this one. I do not understand why.

Remember that a maxima is a point at which f'(x) changes from positive to negative, and a minima is a point at which f'(x) changes from negative to positive.

But for your derivative, there are no points where this is the case (since the square in the derivative is making everything positive)

So in other words, maxima/minima are necessarily critical numbers, but a critical number is not necessarily a maxima/minima.

Here is a graph of your function for visualization

 

[Zoned]

You can also use second derivative test to check the relative maximum or minimum but it's not feasible to do that in this question.

 

[FUBAR McDangles]

Alright, so -10 is undefined so on both sides you would get an asymptote, right? Things approach -10 but don't actually get there. And since everything is positive because of the squared denominator, the sign never switches so by default there is no extrema.

Alright I think I got this ironed down. Many thanks to the both of you.

 

[Zoned]

Will edit this in a second. Just saw the visualization/second post.

X = -10 will be an asymptote.

A general suggestion - practice making graphs for every problem in Calculus. If you master the art of graphing functions like exp, trig, log, inverse, quadratic etc then you'll have a good command on what to do in problems. At least that is what I followed in my first year of math at college for calc 1,2,3 and diff eq

 

[Therion]

Could use some help on this one:


So for relative extrema you find the first derivative. Through the quotient rule i figured it is 10/((x+10)^2) - which I am fairly certain is correct.

From there I am confused. For the critical values, is it just -10 because of the potential /0 scenario in the denominator? What about that 10 floating around up top?

Can anybody help me out here? I would really appreciate it.

We generally only consider the domain of the function when finding critical values (and by extension, extrema). So in this case there are no critical values to be found, and no maxima or minima.

 

[0xCA2]

I am assuming that you understand the basic principles of integration. To solve this question, you need to integrate cos x. Since this is an indefinite integral you will get a constant C which is being asked in the question. To get C, just put 5π/6 in the equation that you will after integrating cos x. Now let's come to how to integrate cos x to find coefficients A and B.
There is a half angle equation for cos x.

cos 2x = 2 sq(cos x) - 1.
Substitute cos x in the original question in the form of cos x/2 using the above equation. I am sure that you know how to integrate sq(cos x), use that principle.
Another identity that is useful is
sin 2x = 2cosx sinx

Nope, my teacher is doing the thing where he introduces a concept first under a different name. We haven't touched on integrals, or the principles of integrals, yet. This has to do with "Antiderivatives" .

Thanks for the help anyway, though.

 

[Leezard]

Hey guys, could you help me with questions like this?


Thanks.

differentiate the function (f (x) = A sin(x)^B cos(x)^C + D) and determine the constants by inspection and comparing with f ' (x) = cos(x), then determine other constants with the other condition (f (5π/6) = 9.5).

Also, are you sure its not f (x) = A sin(x)^B + cos(x)^C + D? It's almost the same question, just a little bit easier.

 

[0xCA2]

differentiate the function (f (x) = A sin(x)^B cos(x)^C + D) and determine the constants by inspection and comparing with f ' (x) = cos(x), then determine other constants with the other condition (f (5π/6) = 9.5).

Also, are you sure its not f (x) = A sin(x)^B + cos(x)^C + D? It's almost the same question, just a little bit easier.

What do you mean by this? This was pretty easy to do when you were given f(0), but now it seems like that really doesn't help much.

Also yeah I'm sure, it's multiplying them.

Here's what I got when I took the derivative:
A[(B(sinx)^B-1)(cosx) - (C(cosx)*C-1)(sinx)]

That's where I am so far. I'm just going to move on for a bit and either go to tutoring or ask the TA about it.

 

[Leezard]

What do you mean by this? This was pretty easy to do when you were given f(0), but now it seems like that really doesn't help much.
.

It's the same thing, just instead of replacing x with 0 into the expression, you replace x with 5π/6.

 

[Therion]

Nope, my teacher is doing the thing where he introduces a concept first under a different name. We haven't touched on integrals, or the principles of integrals, yet. This has to do with "Antiderivatives" .

Thanks for the help anyway, though.

If you're supposed to be using antiderivatives, I suppose the question to ask is whether you know an antiderivative of cos x. I would assume that you do, and for the right choices of A, B, and C, that antiderivative should match the f(x) you've been given. Then you can just determine D using the last piece of information you were given, which as Leezard mentioned is no different from when the problem gives you f(0).

 

[MisterLuffy]

f: R -> R f(x) = x/(x^(2) + 1) , for all real numbers x.
I need help with this problem, I have to prove this function that's either one to one or not one to one.

I would think of setting the equation equal to each other, but another way is to graph the function to see if each x values have their own y values. I went to this site here to graph this function. I thought it would be one to one, but I checked on wolframalpha website and it states that it's not one to one. Can anyone explain to me why this function is one to one or not one to one?

 

[Therion]

f: R -> R f(x) = x/(x^(2) + 1) , for all real numbers x.
I need help with this problem, I have to prove this function that's either one to one or not one to one.

I would think of setting the equation equal to each other, but another way is to graph the function to see if each x values have their own y values. I went to this site here to graph this function. I thought it would be one to one, but I checked on wolframalpha website and it states that it's not one to one. Can anyone explain to me why this function is one to one or not one to one?

The function is not one to one. The graph you linked does not pass the horizontal line test. If you wanted to show it without a graph, you could note that f(1/2)=f(2). In fact, you have

f(1/x) = (1/x)/(1+1/x^2) = (1/x)/((x^2+1)/x^2) = x/(x^2+1) = f(x)

for any nonzero value of x, so you can give many examples of repeated values that prevent the function from being one to one.

 

[0xCA2]

It's the same thing, just instead of replacing x with 0 into the expression, you replace x with 5π/6.

If you're supposed to be using antiderivatives, I suppose the question to ask is whether you know an antiderivative of cos x. I would assume that you do, and for the right choices of A, B, and C, that antiderivative should match the f(x) you've been given. Then you can just determine D using the last piece of information you were given, which as Leezard mentioned is no different from when the problem gives you f(0).

I finally did it and I just realized how easy it was. The f(x) would be sin(x) + 9. So A = 1, B = 1, C= 0 and D = 9. I just did another problem given by the math program as a "practice another version" and I got it right within seconds.

I think I have a mental block that makes me assume that something must be hard even though it isn't, and so I ending up making it harder than it is. I can't believe I spent hours on this.

 

[Two Words]

It's been a long time since I did a problem like this, but -10 would be a critical number since f'(-10) does not exist. Critical values are only when f'(x) either equals zero or does not exist, and -10 is only value for which this occurs for your derivative. So that means -10 is the only critical number. However I think -10 is neither a minima nor a maxima.

Don't you also have to take the value of the end point of each side of the X-interval?

 

[MisterLuffy]

The function is not one to one. The graph you linked does not pass the horizontal line test. If you wanted to show it without a graph, you could note that f(1/2)=f(2). In fact, you have

f(1/x) = (1/x)/(1+1/x^2) = (1/x)/((x^2+1)/x^2) = x/(x^2+1) = f(x)

for any nonzero value of x, so you can give many examples of repeated values that prevent the function from being one to one.

Thanks.

 

[RELAYER]

Differential Equations help

Find particular solution of

y'' + 2y' -3y = 1 + xe^x

yc = C1e^x + C2e^(-3x)

The solutions say that yp is of the form
A + Bxe^x + Cx^2e^x

Why does the particular solution contain the C term?

 

[Biggest-Geek-Ever]

Differential Equations help

Find particular solution of

y'' + 2y' -3y = 1 + xe^x

yc = C1e^x + C2e^(-3x)

The solutions say that yp is of the form
A + Bxe^x + Cx^2e^x

Why does the particular solution contain the C term?

When working with undetermined coefficients, the form of your particular solution can be rather finicky. If you started with assuming Y(x) = A+Bx*e^x, you will probably be unable to solve for those coefficients. This is an indicator that you need to "ratchet up" Y(x) by an additional x power, which is where Cx^2*e^x comes from.

 

[Anustart]

Dunno what I'm doing wrong here. Simple math gone awry somewhere I guess.

Wolfram says cos(2x - pi/13) = 0 is 41pi/52 - pi/2

No matter what I do, I come up with 15pi/52 - pi/2

And evidently these aren't the solutions for interval [0, 2pi)

15pi/52, 41pi/52, 67pi/52, 100pi/52. I'm too dumb for this shit. Saw my error in math, went egads, its 93pi/52 instead of 100, but nope, that doesn't = 0 for god knows what fucking reason.

 

[Digital Snake]

Help! I'm doing a take home exam and am having the hardest time with what seems like a simple problem.

It says 'Solve and check for each equation:

x(5x-2)=0 and x^2+2x-8=0

Can someone help me out?

 

[Anustart]

Help! I'm doing a take home exam and am having the hardest time with what seems like a simple problem.

It says 'Solve and check for each equation:

x(5x-2)=0 and x^2+2x-8=0

Can someone help me out?

For x(5x-2)=0 just take each factor and set them to zero.

First is easy, x = 0, then 5x - 2 = 0 -> 5x = 2 -> x = 2/5

You'll have to factor the second one.

(x + 4)(x - 2) = 0 then set each of those = to 0 and solve.

 

[GiantEnemyGoomba]

Dunno what I'm doing wrong here. Simple math gone awry somewhere I guess.

Wolfram says cos(2x - pi/13) = 0 is 41pi/52 - pi/2

No matter what I do, I come up with 15pi/52 - pi/2

And evidently these aren't the solutions for interval [0, 2pi)

15pi/52, 41pi/52, 67pi/52, 100pi/52. I'm too dumb for this shit. Saw my error in math, went egads, its 93pi/52 instead of 100, but nope, that doesn't = 0 for god knows what fucking reason.

I'm pretty sure 41pi/52-pi/2 and 15pi/52-pi/2 are the same thing considering those two are 1pi apart. My trig is a little rusty but I'm almost positive those four x values (15, 41, 67, and 93pi/52) are correct for [0, 2pi]. They all plug into the equation correctly at least.

 

[Anustart]

So I'm doing the trig functions on sum and differences of angles now, which is all fine, can do it rather easy. Part I don't understand is knowing what quadrant the sum or difference of the angles is in?

 

[Kuroyume]

Hey can someone who knows something about statistics help me out here?

Problem #4: Suppose X is a normal random variable with a mean of 11 and a variance of 14. Using R calculate the following:

P(X < 0)

sqrt(14)
pnorm(0,11.0,3.741657) (#Answer = 0.001641729)

P(|X-10| < 2)

10-2 = 8
10+2 = 12
pnorm(12,11.0,3.741657)-pnorm(8,11.0,3.741657)(#Answer = 0.394027)

Find the 70th percentile of X - 2

qnorm(.70,9,3.464102)(#Answer = 10.81658)

Show your R work for credit. Also, you may have to do some operations on X to be able to use the appropriate R functions. Show these operations.

Can someone look at my solutions and confirm whether I got them right or not? The one I'm really confused most by is the last one "Find the 70th percentile of X - 2" so if someone could provide me help that would be helpful. Thanks.

 

[RELAYER]

Needing help with basic 3-D stuff

I've got most of it but the format in which Webwork wants the answer is unknown to me.

The question

Part A find an equation of the sphere with center (-2,-1,2) with radius 4. Be sure that your formulae are monic.

I put
(x+2)^2 + (y+1)^2 + (z-2)^2 -2 = 0

(The "= 0" part is set, so I just moved the 2 to the other side)
Anyways, it didn't like it and says it's wrong. I'm guessing my formula isn't monic but I have no idea what that means.

Same deal with part B
What is the intersection of this sphere with the xz-plane.

That would be a circle when y = 0.

So
(x+2)^2 + (z-2)^2 -1 = 0

Didn't like that either.

Anyone know what is happening?

 

[Granadier]

Calc 1 test tomorrow.

Pants have been shat for the past week. Hoping this ends up going well.

 

[simplayer]

Why is your r^2 only 2? If the radius is 4, shouldn't you have ...=16?

 

[Biggest-Geek-Ever]

Needing help with basic 3-D stuff

I've got most of it but the format in which Webwork wants the answer is unknown to me.

The question

Part A find an equation of the sphere with center (-2,-1,2) with radius 4. Be sure that your formulae are monic.

I put
(x+2)^2 + (y+1)^2 + (z-2)^2 -2 = 0

(The "= 0" part is set, so I just moved the 2 to the other side)
Anyways, it didn't like it and says it's wrong. I'm guessing my formula isn't monic but I have no idea what that means.

Same deal with part B
What is the intersection of this sphere with the xz-plane.

That would be a circle when y = 0.

So
(x+2)^2 + (z-2)^2 -1 = 0

Didn't like that either.

Anyone know what is happening?

If a polynomial is monic, that means the leading coefficient is 1, so something like x^3+3x^2-2x+1 = 0. I've never seen a multivariable equation referred to as monic, but I suspect they want you to expand out the equation and then have x^2 be your leading term. Also with a radius 4, it should be 16, not 2.

For Part B, did you try adding the 1 over to the right side?

 

[Digital Snake]

For x(5x-2)=0 just take each factor and set them to zero.

First is easy, x = 0, then 5x - 2 = 0 -> 5x = 2 -> x = 2/5

You'll have to factor the second one.

(x + 4)(x - 2) = 0 then set each of those = to 0 and solve.

Thank you very much, I really appreciate it!

 

[RELAYER]

Why is your r^2 only 2? If the radius is 4, shouldn't you have ...=16?

If a polynomial is monic, that means the leading coefficient is 1, so something like x^3+3x^2-2x+1 = 0. I've never seen a multivariable equation referred to as monic, but I suspect they want you to expand out the equation and then have x^2 be your leading term. Also with a radius 4, it should be 16, not 2.

For Part B, did you try adding the 1 over to the right side?

Thanks to both, the radius thing was the issue. Went the wrong way, took the root instead of the square. Derp.

 

[RELAYER]

Is there any point to differential equations other than giving people carpal tunnel syndrome from taking the 4th derivative of an equation that is 5 lines long, then plugging them all back in to a super equation that literally takes up half the page to find some useless coefficients?
I mean come on, really. This time would be better spent showing us how to operate computer programs that do this shit for us without killing half the rainforest to make enough paper and pencils.

After struggling to staple together the packet of sheets I just created, I can at least remind myself that my 14 hours of effort are worth 0.1 percent of my final grade.
Fuck.

 

[MisterLuffy]

Hello. I have a question I need help on. Or more like I need a better understanding on how to do this problem.

Find the big O estimate for functions f(x), x&#8712;R
f(x) = 4x^(3) + x^(2)logx

 

[Stumpokapow]

Hello. I have a question I need help on. Or more like a better understanding on how to do this problem.

Find the big O estimate for functions f(x), x&#8712;R
f(x) = 4x^(3) + x^(2)logx

First, let's ask what "Big O" means? Big O is a function designed to mirror the growth pattern of the function, so we can compare it to other functions with comparable growth patterns. We do this because we want to choose functions with lower Big O to be more efficient, especially when we're dealing with problems that have large n of items, where slower functions can grow very quickly. So, again, it's about finding a function with a comparable growth pattern. With Big O, we take the dominant term.

We do this because at sufficiently large values of x, the full equation will approach the dominant term.

If you imagine the function:
f(x) = x^2 + 2x

At x=2, the function's value is 8, x^2=4, and 2x=4
At x=10, the function's value is 120, x^2=100 and 2x=20
At x=100, the function's value is 10200, x^2=1000 and 2x=200
At x=10000, the function's value is 100,020,000, x^2=100,000,000 and 2x=20,000
Do you see how the dominant term increases much more quickly than the other term?

So we would say the O(x) = x^2 for this equation, because the equation basically grows at the speed of the dominant portion of the equation. If f(x) was 10x^2 + 2x, O would still be x^2, because even though there's a constant term 10 in front of the equation, at high enough x, it's still growing at a rate comparable to x^2, right? You can graph x^2 versus 10x^2 to see that the functions have very similar growth patterns.

Red = x^2 + 2x
Blue = x^2
Green = 2x

Back to your equation, what would you say is the dominant term?
4x^3 or x^2log(x)?

Purple = 4x^3 + x^2log(x) -- you can't even see purple because it's so close to orange.
Orange = 4x^3
Black = x^2log(x)

At x=10:
4x^3 = 4000
x^2log(X) = 200

At x=1000:
4x^3 = 4,000,000,000
x^2log(x) = 3,000,000

It seems to me like the 4x^3 term is the dominant term and we'd call this equation O(x) = x^3


Before you click the below link, maybe stop to consider how certain types of functions might be dominant over certain other types of functions. For example, which is dominant? 2^x or x^2? Which is dominant? x^5 or x^4? Which is dominant? x^2 or xlog(x)?

Here are some rules for determining dominance:
http://ellard.org/dan/www/Q-97/HTML/root/node7.html

 

[Two Words]

C -> log -> sqrt -> n -> nlog -> n^C -> C^n -> n! -> n^n


I believe this is the correct order of growing functions, where n is an infinitely growing variable and C is a constant. I wonder why n^n is ignored in my book and classes though.

 

[MisterLuffy]

First, let's ask what "Big O" means? Big O is a function designed to mirror the growth pattern of the function, so we can compare it to other functions with comparable growth patterns. We do this because we want to choose functions with lower Big O to be more efficient, especially when we're dealing with problems that have large n of items, where slower functions can grow very quickly. So, again, it's about finding a function with a comparable growth pattern. With Big O, we take the dominant term.

We do this because at sufficiently large values of x, the full equation will approach the dominant term.

If you imagine the function:
f(x) = x^2 + 2x

At x=2, the function's value is 8, x^2=4, and 2x=4
At x=10, the function's value is 120, x^2=100 and 2x=20
At x=100, the function's value is 10200, x^2=1000 and 2x=200
At x=10000, the function's value is 100,020,000, x^2=100,000,000 and 2x=20,000
Do you see how the dominant term increases much more quickly than the other term?

So we would say the O(x) = x^2 for this equation, because the equation basically grows at the speed of the dominant portion of the equation. If f(x) was 10x^2 + 2x, O would still be x^2, because even though there's a constant term 10 in front of the equation, at high enough x, it's still growing at a rate comparable to x^2, right? You can graph x^2 versus 10x^2 to see that the functions have very similar growth patterns.

Red = x^2 + 2x
Blue = x^2
Green = 2x

Back to your equation, what would you say is the dominant term?
4x^3 or x^2log(x)?

Purple = 4x^3 + x^2log(x) -- you can't even see purple because it's so close to orange.
Orange = 4x^3
Black = x^2log(x)

At x=10:
4x^3 = 4000
x^2log(X) = 200

At x=1000:
4x^3 = 4,000,000,000
x^2log(x) = 3,000,000

It seems to me like the 4x^3 term is the dominant term and we'd call this equation O(x) = x^3


Before you click the below link, maybe stop to consider how certain types of functions might be dominant over certain other types of functions. For example, which is dominant? 2^x or x^2? Which is dominant? x^5 or x^4? Which is dominant? x^2 or xlog(x)?

Here are some rules for determining dominance:
http://ellard.org/dan/www/Q-97/HTML/root/node7.html

It makes sense when graphing the function and trying to compare two terms. Also, substituting x with an integer to see which answer is bigger than the other. Answering the last question.For the 2^x and x^2, I would use a graphing calculator to compare the graphs and the y values. 2^x is the dominant term. Next, between x^5 and x^4, x^5 is dominant term. Last, x^2 and xlog(x), x^2 is dominant term.

I have another problem I want to do it with you.
f(x) = (x^(5)+ 3x^(2) +2)/(x^(3) + 6)
So we'll start with the numerator, x^5 is the dominant term. For the denominator, x^3 is the dominant term. Therefore x^(5) / x^(3) = x^(5-3) = x^2
O(x) = x^2
Did I do this right?

 

[Two Words]

You could think of that problem as O(f(x))/O(g(x)), which gets you x^2.

 

[Stumpokapow]

Yep, O(x^2) is the order of that function, no matter which way you do it.

One really important thing is that you make sure to expand where possible, EG: f(x) = (x^2 + 6)(x - 3) is NOT O(x^2), because if you expand the terms you get an x^3 term, right?

 

[MisterLuffy]

Yep, O(x^2) is the order of that function, no matter which way you do it.

One really important thing is that you make sure to expand where possible, EG: f(x) = (x^2 + 6)(x - 3) is NOT O(x^2), because if you expand the terms you get an x^3 term, right?

How did you get to that function?

Edit: My bad thanks to Two Words for pointing out that it's an example. Your right, if you multiply them out it would turn out to be (x^3 - 3x^2 + 6x - 18) which leaves us x^3 as the dominant term. Thanks for your help.

 

[Two Words]

It's just an example

 

[MisterLuffy]

It's just an example

oh lol.

 

[spelltropy]

Trying to prove the following:

For some non-empty set A which is bounded above, given e > 0, there exists an element of A, a, such that sup A - e < a <= sup A

Can anyone give me a shove in the right direction? I have a feeling its very straight forward but I'm just not sure what I should be doing...

 

[Therion]

Trying to prove the following:

For some non-empty set A which is bounded above, given e > 0, there exists an element of A, a, such that sup A - e < a <= sup A

Can anyone give me a shove in the right direction? I have a feeling its very straight forward but I'm just not sure what I should be doing...

Try proof by contradiction. Assume that there is no element of A that lies between sup A - e and sup A. Then compare this with the way you defined sup A in the first place.

 

[iSnakeTk]

I have a question .

I need to prove this inequality but have no idea how to even start.


Let G be a graph.

if diam(G)>=4 then diam(G(complement))<=2.

 

[Therion]

I have a question .

I need to prove this inequality but have no idea how to even start.


Let G be a graph.

if diam(G)>=4 then diam(G(complement))<=2.

Edit: Sorry, I misread the question. This is probably true in general after all.

 

[iSnakeTk]

I think you're missing part of the question, as this is not true for arbitrary graphs.

but thats the entire question. I did not leave anything out .

can u show me why that statement is not true ?

 

[Therion]

I have a question .

I need to prove this inequality but have no idea how to even start.


Let G be a graph.

if diam(G)>=4 then diam(G(complement))<=2.

It is easy to show that if diam(G) is infinite, then diam(G(complement))<=2, so consider the case where G has a geodesic of length 4, say A-B-C-D-E. Note that no other edges on this set of vertices can be in G, so they are all in the complement. If G has no other points you are done. If it has another point X, then X can only connect to vertices that are at most two apart.from each other, in which case there will be paths of length at most 2 from X to any vertex in the complement. If there is another vertex Y in G, then you can check that d(X,Y)<=2 in the complement as well by considering how X and Y can connect to each other and the five geodesic vertices in G.

There is probably some clever way of doing all this without an explicit construction, but it's been a while since I've done any graph theory and I don't recall what theorems may be available..