All problems in one post :P (5 questions)

[Towelie]

Q1.If an equilateral triangle and a square have the same area, what is the ratio of their perimeters?

Q2.what is the largest  5 digit palindrome with no digit represented more than twice?

Q3.To the nearest 100th of a square inch, what is the area inside the three point zone but outside the paint in college basketball?

Q4. Under what conditions does a 5th degree polynomial have exactly two inflection points?

Q5.what is the area of the largest isosceles trapezoid that can be inscribed under the sine curve from 0 to pi ?

[d&q]

1. fourth root-3 / six?
2. 98789
3. dont want to look up measurements
4. at least 3 real roots, and two of the roots are equal? [Edit]: you need an odd number, you cant have two points of inflection for a 5th degree polynomial.

[Ender]

(4)

Given the polynomial x^5 + ax^4 + bx^3 + cx^2 + dx + e, it has exactly two inflection points if 2a^2 > 5b.

Hint: An nth degree polynomial has n real roots iff

.

Take this as a given. It's a result of the Cauchy-Schwarz Inequality.

By the way, this is a beautiful problem =)

[Ender]

4. at least 3 real roots, and two of the roots are equal? [Edit]: you need an odd number, you cant have two points of inflection for a 5th degree polynomial.

I'm pretty sure that's wrong.

[Towelie]

(4)

Given the polynomial x^5 + ax^4 + bx^3 + cx^2 + dx + e, it has exactly two inflection points if 2a^2 > 5b.

Hint: An nth degree polynomial has n real roots iff

.

Take this as a given. It's a result of the Cauchy-Schwarz Inequality.

By the way, this is a beautiful problem =)

Rofl, I sent that answer to him and I got this reply:
That sounds pretty smart but I doubt it.  I'm looking for a more general solution.... remember I don't have a caculator & I won't bother looking up solutions on the net.  I waste all my time on the ladies!

[Sidoh]

That sounds pretty smart but I doubt it.  I'm looking for a more general solution.... remember I don't have a caculator & I won't bother looking up solutions on the net.  I waste all my time on the ladies!

What a giant douchebag.

[Ender]

(4)

Given the polynomial x^5 + ax^4 + bx^3 + cx^2 + dx + e, it has exactly two inflection points if 2a^2 > 5b.

Hint: An nth degree polynomial has n real roots iff

.

Take this as a given. It's a result of the Cauchy-Schwarz Inequality.

By the way, this is a beautiful problem =)

Hmm... I take this back. No fifth degree polynomial can have exactly two inflection points. This is because its second derivative can't have two real roots (each with a multiplicity of 1). It can either have three real roots or one real root and two complex roots. And in a cubic function the number of real roots you have corresponds to the number of points where the polynomial changes sign.

The way I got my first solution was by solving for when the fifth degree polynomial and its second derivative satisfy

.

The problem is that when the second derivative has three real roots, it has three inflection points, not two.

The reason I said this problem was beautiful is that it worked out so that if you have a fifth degree polynomial with five real roots, then its second derivative will have three real roots. I suspect that a the ith derivative of an nth degree polynomial with n real roots has n - i real roots, but I'm going to get around to proving it in another thread.

[Towelie]

(4)

Given the polynomial x^5 + ax^4 + bx^3 + cx^2 + dx + e, it has exactly two inflection points if 2a^2 > 5b.

Hint: An nth degree polynomial has n real roots iff

.

Take this as a given. It's a result of the Cauchy-Schwarz Inequality.

By the way, this is a beautiful problem =)

Hmm... I take this back. No fifth degree polynomial can have exactly two inflection points. This is because its second derivative can't have two real roots (each with a multiplicity of 1). It can either have three real roots or one real root and two complex roots. And in a cubic function the number of real roots you have corresponds to the number of points where the polynomial changes sign.

The way I got my first solution was by solving for when the fifth degree polynomial and its second derivative satisfy

.

The problem is that when the second derivative has three real roots, it has three inflection points, not two.

The reason I said this problem was beautiful is that it worked out so that if you have a fifth degree polynomial with five real roots, then its second derivative will have three real roots. I suspect that a the ith derivative of an nth degree polynomial with n real roots has n - i real roots, but I'm going to get around to proving it in another thread.

x^5+x^4+2 does. F'(x)= 5x^4+4x^3, f"(x)= 20x^3+12x^2 which factors into 4x^2(5x+3), inflection points at 0 and -(3/5)

[dark_drake]

x^5+x^4+2 does. F'(x)= 5x^4+4x^3, f"(x)= 20x^3+12x^2 which factors into 4x^2(5x+3), inflection points at 0 and -(3/5)

It really doesn't look that way when I graph it.  :-\  Also, all you proved was that the second derivative had two solutions.  Yes, it is necessary for an inflection point to have f''(x)=0, but it doesn't guarantee that the point is an inflection point.

[Ender]

(4)

Given the polynomial x^5 + ax^4 + bx^3 + cx^2 + dx + e, it has exactly two inflection points if 2a^2 > 5b.

Hint: An nth degree polynomial has n real roots iff

.

Take this as a given. It's a result of the Cauchy-Schwarz Inequality.

By the way, this is a beautiful problem =)

Hmm... I take this back. No fifth degree polynomial can have exactly two inflection points. This is because its second derivative can't have two real roots (each with a multiplicity of 1). It can either have three real roots or one real root and two complex roots. And in a cubic function the number of real roots you have corresponds to the number of points where the polynomial changes sign.

The way I got my first solution was by solving for when the fifth degree polynomial and its second derivative satisfy

.

The problem is that when the second derivative has three real roots, it has three inflection points, not two.

The reason I said this problem was beautiful is that it worked out so that if you have a fifth degree polynomial with five real roots, then its second derivative will have three real roots. I suspect that a the ith derivative of an nth degree polynomial with n real roots has n - i real roots, but I'm going to get around to proving it in another thread.

x^5+x^4+2 does. F'(x)= 5x^4+4x^3, f"(x)= 20x^3+12x^2 which factors into 4x^2(5x+3), inflection points at 0 and -(3/5)

The root x=0 has a multiplicity of two, meaning it doesn't change sign at x=0. As dark_drake said, for a point x=c to be an inflection point in f(x), f''(x) must change sign at x=c. If f''(c) is zero, it doesn't necessarily change sign (e.g., x=0 on a parabola).

[d&q]

4. at least 3 real roots, and two of the roots are equal? [Edit]: you need an odd number, you cant have two points of inflection for a 5th degree polynomial.

I'm pretty sure that's wrong.

Hmm... I take this back. No fifth degree polynomial can have exactly two inflection points. This is because its second derivative can't have two real roots (each with a multiplicity of 1). It can either have three real roots or one real root and two complex roots. And in a cubic function the number of real roots you have corresponds to the number of points where the polynomial changes sign.

The way I got my first solution was by solving for when the fifth degree polynomial and its second derivative satisfy

.

The problem is that when the second derivative has three real roots, it has three inflection points, not two.

The reason I said this problem was beautiful is that it worked out so that if you have a fifth degree polynomial with five real roots, then its second derivative will have three real roots. I suspect that a the ith derivative of an nth degree polynomial with n real roots has n - i real roots, but I'm going to get around to proving it in another thread.

What the hell?

[Ender]

4. at least 3 real roots, and two of the roots are equal? [Edit]: you need an odd number, you cant have two points of inflection for a 5th degree polynomial.

I'm pretty sure that's wrong.

Hmm... I take this back. No fifth degree polynomial can have exactly two inflection points. This is because its second derivative can't have two real roots (each with a multiplicity of 1). It can either have three real roots or one real root and two complex roots. And in a cubic function the number of real roots you have corresponds to the number of points where the polynomial changes sign.

The way I got my first solution was by solving for when the fifth degree polynomial and its second derivative satisfy

.

The problem is that when the second derivative has three real roots, it has three inflection points, not two.

The reason I said this problem was beautiful is that it worked out so that if you have a fifth degree polynomial with five real roots, then its second derivative will have three real roots. I suspect that a the ith derivative of an nth degree polynomial with n real roots has n - i real roots, but I'm going to get around to proving it in another thread.

What the hell?

Oh sorry, Deuce, I missed that you said that. My bad =\ And good job =).

[Towelie]

anyone going to try #5?

[Ender]

I got an area of 1.82 for Q5.

Solution soon to come.

[Ender]

We know that the area of an isosceles trapezoid is the average of the bases times the height. We want to get this area in terms of one variable, x. I chose for x the length of the shorter base. We know that the other base will be pi. Now we need to find out the height that the shorter base is at. The distance from the origin to the shorter base is (pi - x) / 2. The height is the sine of this.



Then you differentiate this function and solve for the roots of the derivative. I did this with my TI-89. There are infinitely many roots, as it contains a trig function, but only one root, .92, is under pi (it has to be, since it's the shorter base). I plugged this root into the area function and got 1.82.

Edit: nvm.~!~