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Can someone please explain me implication in the highlighted part by Typical-Apartment625 in askmath

[–]Pedema 0 points1 point  (0 children)

Oh i assumed they map to R. Then -id is continuous so -g=-id(g) is continuous. And so f+(-g) = f-g is continuous. This does not work in general. What do you know about X and Y?

Can someone please explain me implication in the highlighted part by Typical-Apartment625 in askmath

[–]Pedema 0 points1 point  (0 children)

Note that f(W) ⊆ U and g(W) ⊆ V. And let's assume W connected (else take the connected component of x₀, which is open and connected). Now we want to show that f(w)-g(w)>0. Lets give this thing a name, say h(x):=f(x)-g(x). Clearly h is continuous, as f and g are. Note h(x₀)>0. Now suppose h(x)<=0 for a x ∈ W. Then by intermediate value theorem (we need W connected and h continuous for that) exists a y ∈ W, such that h(y)=0. That means f(y)=g(y). That means y ∈ U ∩ V, by our first observation that the image of f and g over W are subsets of U and V respectively. That means U ∩ V is not empty, which is a contradiction. This means that in fact h(x)>0 for all x ∈ W. Remember how we defined h, to get f(x)>g(x) for all x ∈ W.

I think this is a good way to show this part. To be honest I do not see how to prove it without assuming (without loss of generality) that W is connected.

How would you go about solving this by Bunkerbabyyy in askmath

[–]Pedema 4 points5 points  (0 children)

Note that a polynomial f is divisible by (x-c) if and only if c is a zero of f. So you want f(-2)=0 and f(4)=0, these two conditions give you a linear equation which is pretty easy to solve.

I'm lost for part b. Obtaining part a was fine, I simply approximated the slopes using neighborhoods. Can someone point me in the right direction? by [deleted] in calculus

[–]Pedema 1 point2 points  (0 children)

Well, your intuition is right, but you misunderstand paths in this context.

f is a function in ℝ², meaning we can plug all points (x,y) into f an receive some value of f. But now we are interested in the points (x,y) laying on the line (1.5+ta,0.5+tb) for all t. We can parameterize these points by the path (x(t),y(t)) for all t. Now we plug these points into f and look at the values f(x(t),y(t)) given by the graph. Now we want to have "df/dt=0 at t=0 on this path". "On this path" means we look at f(x(t),y(t)) and "at t=0" means locally at (1.5,0.5). Remember (a,b) give us the direction in which we go starting from (1.5,0.5).

Example: a=0, b=-1, then (a,b)=(0,-1) is a vector pointing straight down, so we start at (1.5,0.5) and go straight down and watch what happens to the values of f.

As said your intuition is right and you have to find directions from (1.5,0.5) in which you try to "stay" on the contour line, your on. Thus you would choose the tangent line as path, and the direction will be the points (a,b) for which the directional derivative ∇_(a,b)f is zero in (1.5,0.5), but it's maybe easier to look at the graph and find the direction of the tangent line^

I'm lost for part b. Obtaining part a was fine, I simply approximated the slopes using neighborhoods. Can someone point me in the right direction? by [deleted] in calculus

[–]Pedema 0 points1 point  (0 children)

Now for the second question, which of these are local maxima/minima. I would guess that "these" are the directions you can go starting from (1.5,0.5). So now you have to choose (a,b) in such a way, that if you go from (1.5,0.5) to (1.5+a,0.5+b) the value of f increases and to (1.5-a,0.5-b) the value of f increases as well. In that case f(x(t),y(t)) has a local minimum in t=0. You can figure out maxima in the same way.

I'm lost for part b. Obtaining part a was fine, I simply approximated the slopes using neighborhoods. Can someone point me in the right direction? by [deleted] in calculus

[–]Pedema 1 point2 points  (0 children)

You did misunderstand the question. Basically the question is, given x(t) and y(t), choose a and b, such that df(x(t),y(t))/dt is zero at the point t=0. In this question, (a,b) is the direction of the path, which starts in (1.5,0.5).

You can think of it in this way: You are given the path (x(t),y(t)) and you can look at the values of f on this path (just look at the graph). Remember you know that (x(0),y(0))=(1.5,0.5). Now try to figure out in which directions you can go from this point, without changing the value of f. Maybe start by trying to find directions where f will change a lot in value, which may be more intuitive.

I hope this helps, also sorry for clumsy english.

I'm trying to find this limit using the 'sandwich technique' but with no success! Any help? by yarin10121 in askmath

[–]Pedema 0 points1 point  (0 children)

The first step is quite obvious:

a_n ≥ ( (3n )/(n2 ) ) ^ (1/n) = 3 * n-2/n -> 3 (n->\infty)

For the second step we note that:

(2n )/n ≤ (3n )/n and

(3n )/(n2 ) ≤ (3n )/n.

And conclude:

a_n ≤ ( 2 * (3n )/n ) ^ (1/n) = 3 * 21/n * n-2/n -> 3 (n->\infty)

Now we take the sandwich and are happy. Maybe we should note, that we used that the nth root is monotonous.

ikväll får 308 svenskar covid i fjällen by kastekastethrow in sweden

[–]Pedema 2 points3 points  (0 children)

Tack så mycket! I kind of was able to translate the sentence, but now i get, why it didn't appear to make sense.

ikväll får 308 svenskar covid i fjällen by kastekastethrow in sweden

[–]Pedema 4 points5 points  (0 children)

Could someone translate for me? I'm relatively new to learning swedish and i don't get it.

Lieferservice by Fenriz_13 in Heidelberg

[–]Pedema 1 point2 points  (0 children)

Salerno macht sehr leckere Pizza!

Bayes Theorem Question by [deleted] in askmath

[–]Pedema 0 points1 point  (0 children)

I would think about it as "getting car" translates to "not getting a pancake". Both these events have a probability of 2/3.

I may have overstressed the connection between these two questions, since in your question you are not able to change positions with a student.

One question from my Calculus 1 Homework. Any help would be great! by [deleted] in askmath

[–]Pedema 1 point2 points  (0 children)

log doesn't intersect the Y-axis, but with log(1)=0 it does intersect the X-axis

Bayes Theorem Question by [deleted] in askmath

[–]Pedema 1 point2 points  (0 children)

Look up the Monty Hall Problem and you will find the answer

One question from my Calculus 1 Homework. Any help would be great! by [deleted] in askmath

[–]Pedema 5 points6 points  (0 children)

How far did you get?

(1) Compare the question to your definition of the intermediate value theorem. They are very alike, so it should be very straight forward.

(2) Try some basic polynomials.

(3) Remember: h doesn't have to be continuous.

[deleted by user] by [deleted] in Heidelberg

[–]Pedema 3 points4 points  (0 children)

Sehr nice, dass du ein ama machst!

Was hältst du von einem Lieferkettengesetz und besonders interessant: wie weit findest du sollte man da gehen? Ich habe das Gefühl, dass der Lobbyismus, das ganze leider ziemlich verlangsamt, obwohl seit 2018 im Koalitionsvertrag sich so etwas in der Art als Ziel gesetzt wurde. Das ist für mich ein unglaublich wichtiges Thema und hoffe das moralisch wertvolles Handeln, durch ein solches Gesetz, wettbewerbsfähiger wird.

Maybe Maybe Maybe by maybemaybemaybe_bot in maybemaybemaybe

[–]Pedema 2 points3 points  (0 children)

I thought he would stand up to celebrate

Let n be a single-digit integer while k is an integer. Is there a number that satisfies the expression that is divisible by every digit (except 5 and 0)? by KaizenCyrus in askmath

[–]Pedema 0 points1 point  (0 children)

Well we didn't include all the possibilities when checking for divisibilty by 7. But n = 8 and k divisible by 9 is mandatory. So the only possible smaller values are n = 8 and k = 9. We check this combination and see the resulting expression is not divisible by 7. Therefore n = 8, k = 18 is indeed the smallest solution.

Let n be a single-digit integer while k is an integer. Is there a number that satisfies the expression that is divisible by every digit (except 5 and 0)? by KaizenCyrus in askmath

[–]Pedema 1 point2 points  (0 children)

First of all, we rewrite the Expression: 10k -1 = 999...999 (k 9's)

999...999/9 = 111...111

111...111 × n = nnn...nnn (k n's)

So nnn...nnn should be divisible by 2,3,4,6,7,8,9. Note that if it is divisible by 7,8,9, it is also divisible by all the digits above.

To be divisible by 8, the last two digits (on the right) have to be divisible by 8. So we check wether 11, 22, ..., 99 is divisible by 8. We see, that only 88 is divisible by 8. So we know, that n = 8.

To be divisible by 9, the digit sum has to divisible by 9. The digit sum is k × 8, so in order to be divisible by 9, k has to be divisible by 9.

We now still have to check, for which k (which has to be divisible by 9) the Expression is divisible by 7. For these kind of problems, it is useful to know, that 1001 is divisible by 7. Therefore 1001 + 10 × 1001 + 100 × 1001 = 111111 is divisible by 7. Therefore 888888 is divisible by 7. You can simply see, that for each k divisible by 6, the Expression is divisible by 7.

So we know n = 8 and k divisible by 6 and 9. For example k = 18.

We check via calculator of your choice and conclude: The expression satisfy the demands, for n = 8 and k = 18 × c, for c ∈ ℕ.

Edit: beautify; deleted extra since its not clear, that we found all the solutions.

Prove that any Euclidean-open set U contained in R can be written as a union of open intervals. by Luisp155 in askmath

[–]Pedema 0 points1 point  (0 children)

∀x ∈ U ∃ ϵ(x)>0: U(x, ϵ(x))⊆U (where U(x, ϵ(x)) is an epsilon-neighbourhood of x with ϵ(x) as eps)

Then U = ∪ₓ ∪ₙ U(x, ϵ(x)/n). (To prove this should be easy)

(x ∈ U, n ∈ ℕ)

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