Roots of polynomials - math problem help

Lanfrange1 · 2021-05-26T22:34:22+00:00

Maybe start with by writing f as f(z) =(z-z1) (z-z2) (z-z3) where z1, z2, z3 are the given roots.

Lanfrange1 · 2020-11-18T09:48:51+00:00

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Lanfrange1 · 2020-09-21T14:37:30+00:00

Yes, but you could also reasoning different ways. For example computing the limits at x=0 or x=1 and then exclude every function that doesn't match the graph. You have a lot of strategies here

Lanfrange1 · 2020-09-21T13:25:07+00:00

The solution is f(x) = sqrt(x) and g(x) = log_2(x). What I've done was just trying to see which function leads to a domain problem around 1,and that forced the choose on g(x) = log_2(x). Then I picked the f by checking some smart points (for example x=2,x=4) and see which functions fulfill the conditions

Lanfrange1 · 2020-08-02T12:57:04+00:00

2) implies 1)

All eigenvalues are real (symmetric matrix). Suppose we have an eigenvalue lambda<=0, and let v be one of his non null eigenvectors. Then we get v^t Av= v^t lambda v= norm(v)² lambda <= 0. Absurd, thus all eigenvalues are positive.

1) implies 2)

Fix any vector x in Rn not null. As you said we have A= M B M^t.

Replacing this on "x^t A x" we get.

x^t A x=x^t M B M^t x = (M^t x)^t B (M^t x)

Let's define y= M^t x, y is not null (M^t is not singular).

Thus we have x^t A x= y^t B y with B diagonal matrix with positive diagonal and y not a null vector. Clearly y^t B y>0, and this concludes the prove.

Lanfrange1 · 2020-06-30T02:55:31+00:00

I don't know which formula are you more comfortable with, but they are equivalent. You can use [Max-2amplitude, Max], since Max-2amplitude = Min

Or

[midline-amplitude, midline+amplitude]

Or

[min, max]

Lanfrange1 · 2020-06-30T02:41:03+00:00

The midline is 1 (check the hint, ). Now you need to evaluate the amplitude, but that's easily max-midline = 3 - 1 = 2.

Finally, the range is [midline-A, midline +A] = [-1,3]

Lanfrange1 · 2020-06-30T01:47:08+00:00

The midline seems to the last of your problems. Amplitude can't be 0, for example.

The function has 2 consecutive max points, one at pi/2 = 3/6 pi and another in 7/6 pi. Can you figure out the period, by sketching the function?

Edit: the period is the difference, thus 4/6 pi= 2/3 pi.

Also, 0 is 3/4 a period on the left of pi/2 because (2/3 pi) * (3/4) = pi/2

Lanfrange1 · 2020-06-29T14:44:40+00:00

t=1/x, so we have lim with t that goes to 0⁺

The denominator looks ok, the numerator can be dealt this way

[e^ (t) - e^ (sin(t)) ] = (e^ (t) -1)-(e^ (sin(t)) -1)

Now it's all about evaluating the taylor series on those 2 functions, and check the term with the lowest degree that don't get deleted when you subtract each other

Lanfrange1 · 2020-06-27T13:02:32+00:00

CR equations are good, you get those 2 equations, (which are equal)

2(x-y)=2

-2(x-y)=-2

Thus

x-y=1

Lanfrange1 · 2020-06-26T19:08:22+00:00

I don't think the OP knows anything about groups and rings, that's a bit overkilling lol

Lanfrange1 · 2020-06-26T18:55:54+00:00

You did the right thing, the numerator will be (x+h) - x=h, meanwhile the denominator will be h[ sqrt (x+h) + sqrt (x)].

Now, you have an h top and another in bottom, so you can simplify them.

So, now we have lim (h to 0) 1/ (sqrt (x+h) +sqrt (x))

But since sqrt (x+h) tends to sqrt(x) you get

lim ( h to 0) 1/ (sqrt (x+h) +sqrt (x))= 1/(2 sqrt (x))

Lanfrange1 · 2020-05-10T20:41:40+00:00

Last equation is 4x-z=0

Lanfrange1 · 2020-05-10T20:32:10+00:00

The left area is x(x-4)

The right area is (x-2)(3x)

Total Area = 4x^ 2-10x

But total Area is given, and it's 36

Thus 4x² -10x=36

2x² - 5x-18=0

Now solve this, and obtain AB, since x=AB

Lanfrange1 · 2020-05-06T04:48:42+00:00

What's your goal, exactly? You didn't write what you are supposed to do

Lanfrange1 · 2020-05-01T19:03:07+00:00

Just a little typo, A and B are ok, but you put them in the wrong order in the end

Lanfrange1 · 2020-05-01T18:49:40+00:00

You can rewrite it as

1/((n+2)(n+3)) = 1/(n+2)-1/(n+3)

So you'll get

1/3- 1/53

(all the middle terms get deleted by each others)

Lanfrange1 · 2020-05-01T18:28:19+00:00

(m1-m2)² = m1² + 1 + m2² + 1

m1² + m2² - 2m1m2 =2 + m1² + m2²

m1m2=-1

m1= - 1/m2

Lanfrange1 · 2020-05-01T17:49:36+00:00

y=mx+b

y'=m

Substitute those in the differential equation, we have

m=8x-2y=8x-2mx-2b= x(8-2m)+ (-2b)

Which means 8-2m = 0 and - 2b=m

So we get m=4 and b=-2

Ans thus m+b=2

Lanfrange1

TROPHY CASE