Partial Gamification
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The mean deviation is also the absolute deviation, I'm casually browsing this.
from a physics forum:
MAD: Mean Absolute Deviation
Good deal, there are a number of approaches.
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The mean deviation is also the absolute deviation, I'm casually browsing this.
from a physics forum:
MAD: Mean Absolute Deviation
Yeah my original thought was to find a normal plane that included the point, then find t to find the point on the vector that was closest to the point and then to do the distance formula but it was getting messy and i couldn't come to the correct answer.
Partial Gamification
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Bumping the orignial question since I was of no real help here. Good luck all.
Thanks. I haven't gotten any further.
youngwerther
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I'm really striking out on derivatives.
Can anyone show me how something like this would be done?
Find f'(a) for f(x) 4/(root(1-x))
Can anyone show me how something like this would be done?
Find f'(a) for f(x) 4/(root(1-x))
First get the 4 out:
4* derivative[(1/sqr(1-x) ] =
= 4 * ((1-x)^-1/2) =
= 4 * (-1/2) [(1-x)^(-1/2 - 1)] (-1) =
= 4 * (1/2 * (1-x)^-3/2) =
= 2 / sqr[(1-x)^3)
Let me know if you need help with any specific step.
youngwerther
Banned
I don't know what you did.
We are supposed to use the
f(x+h)-f(a)/h
formula
The one for babies.
We are supposed to use the
f(x+h)-f(a)/h
formula
The one for babies.
Oh, I don't recall how to do it that way (and you probably mean lim f(x+h)-f(x)/h) , I used derivation rules which is easier; If you're interested, for instance, the derivative of x^2 is 2* x^(2-1)*1: the exponent "comes down" and you decrease one degree and multiply by the derivative of x.
Also 1/x is x^-1 and square root of x is x^1/2
So in your example, 1/sqr(x-1) is the same as having (x-1) to the -1/2 ("-" because we inverted it; and "1/2" because it's the square root. Using the above rule:
[(1 - x)^-1/2)' = (-1/2) (the exponent comes down) * [(1-x)^(-1/2 - 1)] (subtract 1 from your exponent) * -1 (the derivative of (1-x))
Then it's just algebra - in the end you multiply by the four you put aside in the beginning.
Also 1/x is x^-1 and square root of x is x^1/2
So in your example, 1/sqr(x-1) is the same as having (x-1) to the -1/2 ("-" because we inverted it; and "1/2" because it's the square root. Using the above rule:
[(1 - x)^-1/2)' = (-1/2) (the exponent comes down) * [(1-x)^(-1/2 - 1)] (subtract 1 from your exponent) * -1 (the derivative of (1-x))
Then it's just algebra - in the end you multiply by the four you put aside in the beginning.
So I did a quick refresh and like I said, this way it takes more work (not sure if it's for babies). This one was a bit tricky because you have to use several...er, tricks. So you need to replace x by x+h in one of the equations and start solving.
See if you can follow it:
I think it's a little easier (i.e. you'll have fewer fractions to deal with) if you start by getting a common denominator to combine f(x+h)-f(x) into a single fraction. Then rationalize the numerator of that fraction and absorb the h into its denominator. That h will cancel and you can substitute 0 for the remaining h's.
Partial Gamification
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I'm having a really hard time factoring this: 4x^2 - 53x +153 = 0
I realize I can solve this via the quadratic equation, but I'm trying to improve my factoring. So far it hasn't been going very well as you can see.
Look at the constant term's factors: 153 = 17*(3^2)
Try setting up: (4*x + a ) * ( x + b ) = 4*(x^2) + (a+4*b)*x + a*b
When computing absolute and relative error using say, 4-digit chopping/rounding, to how many decimal places does the exact value have to be?
Example:
Abs Error: |p-p*|
e = (the sum of) 1/n! from n=0 to 5 using 4-digit chopping:
=> p* = 1 + 1 + .50 + .1666 + .0416 + .0083 = 2.4665 (Do we have to chop this as well?)
Not using 4-digit chopping, p = 2.716666...
So, do we chop to get 2.716 then plug it into the Abs Error?
Example:
Abs Error: |p-p*|
e = (the sum of) 1/n! from n=0 to 5 using 4-digit chopping:
=> p* = 1 + 1 + .50 + .1666 + .0416 + .0083 = 2.4665 (Do we have to chop this as well?)
Not using 4-digit chopping, p = 2.716666...
So, do we chop to get 2.716 then plug it into the Abs Error?
Partial Gamification
Banned
Are you performing the chop at every operation?
Yes, so:
1/0! = 1
1/1! = 1
1/2! = .5
1/3! = .1666666.. = .1666
1/4! = .0416666... = .0416
1/5! = .0083333... = .0083
Partial Gamification
Banned
Then you got it, four decomal places. the kth chop will be at the four decimal place.
The exact value should be an expression or rational number. Its the exact value. maybe I misunderstood the question.
The exact value should be an expression or rational number. Its the exact value. maybe I misunderstood the question.
I meant that when I compute the Abs error, will it be:
|2.71666.... - 2.4665|
or
|2.7166 - 2.4665|
So, is it beneficial for p (in |p-p*|) to be the true exact value, or can we take the same amount of precision that p* has?
Partial Gamification
Banned
Yeah the true value. Think of the absolute value as a distance, determined by the difference of the exact and approximate values.
youngwerther
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How would one do something like this
let
f(x)= squareroot(1+3x)
find
f'(5)
let
f(x)= squareroot(1+3x)
find
f'(5)
Partial Gamification
Banned
take derivative:
f'(x) = (3)(1/2)*(1+3x)^(-1/2)
plug in value.
edit: remember sqrt(x) = x^(1/2)
youngwerther
Banned
my algebra must suck because i don't know how you got that
the three times the one half, the addition, the negative half power instead of half power.
totally lost
the three times the one half, the addition, the negative half power instead of half power.
totally lost
f(x)=squareroot(1+3x)
It's part of the chain rule for derivatives (derivative of the outside w/ inside unchanged * derivative of the inside)
f'(x)=(1/2)(1+3x)^-1/2^ * (3)
Bold = derivative of the outside (The power comes down in front for the derivative and is reduced by 1 for the power (1/2 - 1 = -1/2); inside remains unchanged)
Italics = derivative of the inside
Partial Gamification
Banned
edit: I like the mention of the chain rule above. Good show!
Derivatives are like reducing the power of an expression, and bringing the original vale to the coefficient.
For a simple equation: f(x) = x^2 , the derivative is: f'(x) = 2*x
The exponent it reduced by one during this process.
note: N(x) = 3*x^4 has the derivative: N'(x)= (3)(4)*x^3 = 12*x^3
If the exponent is one: g(x) = x , then the derivative is: g' (x) = 1 this one is the coefficient
from g(x) = 1* x = x
So, another example: h(x) =2*x^2 + 5*x has the derivative h'(x) = 4*x +5
The derivative of a constant term will be zero: M(x) =5 had a derivative M'(x) =0
With your question, f(x) = sqrt(1+3*x) = (1+3*x)^(1/2)
there are two operations to consider, (1+3x) , and that expression to the 1/2.
the derivative of 1+3x is 3
the derivative of y^(1/2) is (1/2)*y^(-1/2)
So the derivative of (1+3x)^(1/2) = 3*(1/2)*y^(-1/2) but y was that quantity 1+3x
" " = 3(1/2)(1+3x)^(-1/2)
youngwerther
Banned
Oh, we didn't learn the chain rule yet.
He specifically told us we aren't allowed to use it until we cover it in class.
On Friday people used the chain rule and quotient rule on the quiz and got the questions wrong because we hadn't covered it yet.
The lectures seem to be falling behind the homework so everything is a mess to me right now.
He specifically told us we aren't allowed to use it until we cover it in class.
On Friday people used the chain rule and quotient rule on the quiz and got the questions wrong because we hadn't covered it yet.
The lectures seem to be falling behind the homework so everything is a mess to me right now.
youngwerther
Banned
Ok thanks guys, I think I'm learning.
How about this one
f(x)=1/x^2
so the derivative of 1 is zero
x^2 is 2x
so is the derivative 0/2x
?
How about this one
f(x)=1/x^2
so the derivative of 1 is zero
x^2 is 2x
so is the derivative 0/2x
?
Not quite. You can't just take the derivative of the numerator and denominator and put them back where they were.
The easiest way (in my opinion) to do this problem: 1/x^2 equals x^(-2). If you haven't see that idea before, don't be intimidated by it, it's actually fairly simple multiplication. We know 1/x = x^(-1). So 1/(x*x) = (1/x) * (1/x) = (x^(-1))*(x^(-1)) = x^(-2). The lesson here: for any n, 1/x^n = x^(-n).
Once you've grasped that, the derivative is easier to calculate.
f(x)=1/x^2=x^(-2)
f'(x)=(-2)*(x^(-21))
f'(x)=-2(x^(-3))=-2/(x^3)
Technically with this problem, you could do quotient rule. But since it's quite unnecessary and you apparently haven't gotten there yet, that can wait for a while.
youngwerther
Banned
Ok great thanks.
I've got two more to do.
f(x)=x^9h(x)
h(-1)=3
h'(-1)=6
find f'(-1)
So I know we're gonna have f'(x)h(x)+f(x)h'(x)
x^9 would be 9x^8, but when I'm finding the derivative of f(x), what do I do with the h(x)?
I've got two more to do.
f(x)=x^9h(x)
h(-1)=3
h'(-1)=6
find f'(-1)
So I know we're gonna have f'(x)h(x)+f(x)h'(x)
x^9 would be 9x^8, but when I'm finding the derivative of f(x), what do I do with the h(x)?
Is it x^(9h(x)) or (x^9)*(h(x))? Has to be the latter. I'm going to assume the latter.
You have the correct formula with product rule there, but you aren't actually going to see it in that exact form. See, what you actually have here is
f(x) = (x^9)h(x) = g(x)h(x) where g(x)=x^9
So, use that handy product rule:
f'(x) = g'(x)h(x) + g(x)h'(x)
Gonna leave the last part up to you, wanna see if you can get it. I think I've pushed you far enough in the right direction. If not I can type the rest up quick.
youngwerther
Banned
So would it simply be
f'(x) = (9x^8)(3)+(x^9)(6)
f'(x) = (9x^8)(3)+(x^9)(6)
Well,
f'(x) = (9*(x^(9-1)))h(x) + (x^9)h'(x)
f'(x) = 9(x^8)h(x) + (x^9)h'(x)
Is correct. But you can't put the 3 or 6 there yet. That's not true for all x. We only know it's true when x=-1. So you need to make that substitution for every x.
Partial Gamification
Banned
I'll leave a team name suggestion open but here's a pick-up line that is pretty smooth:
"I wish I was your derivative so that I could lay tangent to your curves."
youngwerther
Banned
oh ok, I see.
The last question is about "difference quotients" which is something we haven't covered yet.
It says
Let f(x) = (x^3)-2x
Calculate the difference quotient
(f(3+h)-f(3))/h
and then they give me a bunch of h's, like .1, .01, et cetera.
I recognize the second formula as the one we used to use for calculating derivatives, but other than that, I dunno.
The last question is about "difference quotients" which is something we haven't covered yet.
It says
Let f(x) = (x^3)-2x
Calculate the difference quotient
(f(3+h)-f(3))/h
and then they give me a bunch of h's, like .1, .01, et cetera.
I recognize the second formula as the one we used to use for calculating derivatives, but other than that, I dunno.
They're showing you what happens at h approaches 0, which is the definition of the actual formula for calculating derivatives. It'll get closer and closer to the true "instantaneous slope" of the function at that point.
But there's absolutely nothing hard about this problem; it's Algebra 1. If h is .1, the problem becomes (f(3.1) - f(3)) / 0.1, and so on. Just plug and chug.
ThanksVision
Member
Should I be taking diff eqs before multivariable calc? I don't really know where else to ask. Peers/upperclassmen say I'll be fine but most of the other engineering majors at my school have it so students take multivariable first.
edit: my engineering discipline has it so I take diff eqs first. I don't think the order will matter that much, but I just want to be safe and ask around.
edit: my engineering discipline has it so I take diff eqs first. I don't think the order will matter that much, but I just want to be safe and ask around.
CoffeeJanitor
Member
I took diff EQ and then multivariable.
I honestly don't think it would matter what order you took them in.
Alright I'm working on some linear algebra and am confused on how to prove axioms to show something is a vector space.
Let C be the set of complex numbers. Define addition on C by
(a + bi) + (c + di) = (a + c) + (b + d)i
and define scalar multiplication by
k(a + bi) = ka + kbi
for all real numbers $. Show that C is a vector space with these operation.
I have the list of axioms, but most of them look rather obvious like X + Y = Y + X. How would I prove this exactly?
Let C be the set of complex numbers. Define addition on C by
(a + bi) + (c + di) = (a + c) + (b + d)i
and define scalar multiplication by
k(a + bi) = ka + kbi
for all real numbers $. Show that C is a vector space with these operation.
I have the list of axioms, but most of them look rather obvious like X + Y = Y + X. How would I prove this exactly?
Do you want to prove that C is a R-Vecspace ?
if so for example about the commutativity axiom :
Let u and v be 2 elements of C. Then there are 4 numbers in R, a, b, c and d so that :
u=a+ib and v= c+id
Now u+v = (a+c)+i(b+d) and since the addition is commutative in R you get
u+v=(c+a)+(id+b) = v+u
so the addition in C is commutative
an so on
(writing maths without latex is a shame)
I guess I'm confused with the steps needed. It all seems obvious. For axiom 2 for example I need to prove that (X+Y)+Z = X+(Y+Z) I look at the problem and it looks so simple I'm not sure how to describe the process for proving it.
Quotient rule using Trigonometric Identities:
F(x)=(7-cosx)/(7+sinx)
F'(x)=(sinx)(7+sinx)-[(cosx)(7-cosx)]
F'(x)=7sinx+sin^2(x)-[7cosx-cos^2(x)]
F'(x)=7sinx+sin^2(x)-7cosx+cos^2(x)
Then I dont know what to do...
All of this over (7+sinx)^2 of course.
Edit: Nevermind : sin^2(x)+cos^2(x) =1
F(x)=(7-cosx)/(7+sinx)
F'(x)=(sinx)(7+sinx)-[(cosx)(7-cosx)]
F'(x)=7sinx+sin^2(x)-[7cosx-cos^2(x)]
F'(x)=7sinx+sin^2(x)-7cosx+cos^2(x)
Then I dont know what to do...
All of this over (7+sinx)^2 of course.
Edit: Nevermind : sin^2(x)+cos^2(x) =1
Early exemples in algebra are usually quite straightforward, since everything is based on operations in R it's often only a matter of writing it down properly.
for the associativity rule you need to do "the math" on both sides :
compute (X+Y)+Z and X+(Y+Z). You'll get the same result (a+c+e)+i(b+d+f) and you'll have proven that the addition in C is associative.
Treefingers
Member
Use the fact that real addition is associative, plus the definition of complex addition:
Start with 3 elements of C:
x = a + bi
y = c + di
z = e + fi
where a, b, c, d, e and f are real numbers.
(x + y) + z = [(a + bi) + (c + di)] + (e + fi)
= [(a + c) + (b + d)i] + (e + fi) (using the definition of complex addition)
= [(a + c) + e] + [(b + d) + f]i (again using the definition of complex addition)
= [a + (c + e)] + [b + (d + f)]i (since addition of real numbers is associative)
= (a + bi) + [(c + e) + (d + f)i] (using the definition of complex addition)
= (a + bi) + [(c + di) + (e + fi)] (using the definition of complex addition)
= x + (y + z)
Hopefully this method makes sense, the idea is to express the elements in terms of real and complex parts, then perform complex addition, then move brackets around within the resulting complex element, and then do the reverse. You would follow a similar procedure with the other properties.
Schoolwork, didn't know if I should post this here or in the programming thread. Anyway, relational algebra (SQL):
Basically, I have a table with two columns, A and B that can take on different values. A and B does not have unique values, but two rows cannot be identical. I need to find which A's that "map" to exactly 2 different B's, without functional operators (meaning only set operators and cartesian products).
In this example, x maps to too many Bs, y satisfies the condition and z has too few B's. I can figure out how to find if an A has some, or 1, or no B's, but I have no idea how to narrow it down to a specific number using only logic.
Basically, I have a table with two columns, A and B that can take on different values. A and B does not have unique values, but two rows cannot be identical. I need to find which A's that "map" to exactly 2 different B's, without functional operators (meaning only set operators and cartesian products).
Code:
A | B
x | 2
x | 3
x | 4
y | 2
y | 4
z | 3In this example, x maps to too many Bs, y satisfies the condition and z has too few B's. I can figure out how to find if an A has some, or 1, or no B's, but I have no idea how to narrow it down to a specific number using only logic.
Okay the step by step helps a lot. I think I'm grasping it better.
This homework is just much harder to grasp than previous ones. I have two more types of questions that I'm not sure about. The first is proving a set is a subspace and the other is if the set is spanning.
From what I'm reading, to prove if a set is a subspace you need to show it multiplied by a scalar as well as added to another set and if the requirements are still true its a subspace?
As for spanning, I'm trying to read what it is but I'm not understanding it.
Treefingers
Member
To show that a subset of a vector space over a field K is a subspace you just have to show 3 things:
1. That the 0 element is in the subset
2. That for any elements u and v which belong to the subset, u+v also belongs to the subset
3. That for any element u of the subset and any element k of the field K, ku belongs to the subset.
Just a quick example: C is a vector space over R. The set of elements of C with even real and complex parts, i.e. elements of the form a+bi where a and b are even, is a subspace:
1. 0 is an even number, so 0 belongs to the subset
2. Let u=a+bi, v=c+di belong to our subset. Then a, b, c and d are even. Then u+v = (a+c) + (b+d)i. Since a+c and b+d are even, u+v also belongs to the subset.
3. Let k be a real number and u=a+bi belong to the subset. ku = k(a+bi) = (ka) + (kb)i. Since a and b are even, ka and kb are also even, so ku belongs to the subset.
Hope that helps.
Man you should write a linear book. That's much easier to understand. I should be able to knock those problems out pretty easily now.
I think the only other issue I have now is the last section with vector spans.
smokeandmirrors
Banned
L'Hopitals rule:
Limit as x -> infinity of: (( f(x)) / (g(x))) == ((f ' (x)) / (g ' (x))
In other words, a tricky limit like:
limit as x -> infinity of: ((ln x) / (x))
can be evaluated to:
limit as x -> infinity of ((1/x) / (1))
where
f(x) = ln x
f ' (x) = 1 / x
and
g(x) = x
g ' (x) = 1
so:
limit as x -> infinity of ((1/x) / (1)) = 0, since (1/x)/1 == 1/x, and since x will grow infinitely, 1/infinity is going to be infinitely small (duh) it can be counted as being = 0.
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