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The Math Help Thread
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Note that x and y need not be different
I guess there may be other definitions, but reflexive usually means that (x,x) ∈ R, nothing more.
I hoped it was clear that I was not referring to the reflexive property here but instead giving a hint wrt
antisymmetry
Are you sure? The ratio test is one way to find the radius of convergence, but really you should be looking at your Maclaurin series and asking yourself, "For what values of x does this converge?"
This might sound dumb, but I'm trying to teach myself calculus. I haven't done serious math for about 8 years. There is an example problem in a book I'm reading:
f(x) = x^2 - 9x
f(x) = - x - 9x
For these functions, calculate f(a) and f(x with a little zero at the bottom right)
The answer reads a^2 - 9a; - a^2 - 9a
Which is obvious for the first variation. But what is going on with the second? It has no answer listed. I am also trying to remember what the small 0 designates... Is it a logarithm?
f(x) = x^2 - 9x
f(x) = - x - 9x
For these functions, calculate f(a) and f(x with a little zero at the bottom right)
The answer reads a^2 - 9a; - a^2 - 9a
Which is obvious for the first variation. But what is going on with the second? It has no answer listed. I am also trying to remember what the small 0 designates... Is it a logarithm?
x with the little 0 (x naught) generally represents the starting value for x. so if you're doing physics and have a fuction that represents position of an object, x naught would be the starting x value. is there anything in the example problem that might indicate a starting value?
While it could mean something different in your specific problem, x_0 (read x-naught or x-zero) usually just represents some particular value of x, i.e. a constant rather than a variable. If that's the case here, you can just substitute it into f(x) the same way you did with a, to get essentially the same result.
The goal wasn't to find a value but to develop an understanding of the format of functions. I didn't remember what x naught signified which is why I was confused. I hope I'm able to make it through this book without too much trouble. I forget a lot of trig, which is where I left off 8 years ago.
Thank you for the replies.
Well I'm not sure because I'm still confused about the structure of the Maclaurin series.
I never found the general case of that function.
To get the coefficients I just used the derivatives, evaluated them at a=0, and then divided by the appropriate factorial.
I don't really understand how to get the series yet.
You're being too formulaic. There is no need to use derivatives to get the coefficients or no need to use ratios to get the convergence radius.
The McLaurin series is essentially a polynomial used to represent another function. The function you're trying to represent is already a polynomial. Try to figure it out from there, conceptually.
Rich Uncle Skeleton
Member
Once you know all the coefficients, you know the series. You found the first few. Can you see what all the others are?
I can see that the X is being lowered by one degree each term and that the other coefficients besides the ones I found are just zero.
I just don't see how to express that as a general series. I don't even know if I'm taking the right approach. Sorry for being so dense, I'm just new to these concepts and kind of lost. Is there anything more you could tell me?
I just don't see how to express that as a general series. I don't even know if I'm taking the right approach. Sorry for being so dense, I'm just new to these concepts and kind of lost. Is there anything more you could tell me?
Maybe it'd be helpful to consider that two polynomials are equal only if all their coefficients are equal. With that, it may be easier to understand what the McLaurin representation is.
If you're still lost:
the McLaurin representation and the polynomial are the same function (if you aren't getting why you should try reading the basics again!). That is, c_n = 0 for all n > 3. Then, the "series" is actually just finite and hence the radius of convergence is infinite.
Write everything in terms of cos and sin and it should be easy!
boviscopophobic
Member
The straightforward answer is to express cotangent and tangent in terms of sines and cosines, add the fractions, then simplify. But that is boring, so here is a geometric proof that I whipped up (sorry for crude Paint diagram).
Label the point where the two right angles are marked as E, I forgot to add that label. If we construct two right triangles with a common side BE of length 1 and having angle theta in the marked locations, we see that the side AE is of length tan(theta), and CE is of length cot(theta). The area of triangle ABC is 1/2 is base times its altitude. The altitude BE is length 1, and the base AC is of length tan(theta) + cot(theta) as we just found. Now angle ABC is itself a right angle, so triangle ABC makes up half of the rectangle ABCD. Thus ABCD's area is twice the area of ABC, so it is tan(theta) + cot(theta). On the other hand, the side lengths of ABCD are sec(theta) and csc(theta), so we have that tan(theta) + cot(theta) = sec(theta) csc(theta).
That did the trick, thanks :3 That's actually what I did at first, but forgot the fact that cos^2 + sin^2 = 1.
And I appreciate the visualization too
Need some help with a problem in statistics.
How much do we need to reduce the sample size in order to increase the width of a confidence interval by 50%?
I know it's some sort of algebraic manipulation of a formula, I just can't seem to figure out how to do it. I've been working on it for two hours. Any help, Stats GAF?
How much do we need to reduce the sample size in order to increase the width of a confidence interval by 50%?
I know it's some sort of algebraic manipulation of a formula, I just can't seem to figure out how to do it. I've been working on it for two hours. Any help, Stats GAF?
walking fiend
Member
It depends on your data in fact. Are you assumimg that your variance is the same either for the original sample size or the reduced sample size?
just replace x with the function you are attempting to put into the other function. Whether it is an exponent or not does not matter. 2^(x^3) is one answer. You should be able to find out which one it is and which is the other one without problems.
Oh wow. I was overthinking this a lot. Thanks.
Question: When doing implicit differentiation, how do you know where to put the derivative? I notice that sometimes it's placed next to each term, however when working with a product it placed only next to one term (in the product). Why is this?
For example, from this::
to this:
For example, from this::
to this:
FortuneFaded
Member
You only place y' when you are differentiating y with respect to x. When you are differentiating just a function of x, f(x) becomes f'(x) however when differentiating a product like in your example f(x)g
becomes f'(x)g + f(x)d/dy(g)y'. Effectively you are cancelling the dys. So d/dx(g) becomes d/dy(g)*dy/dx.The dy/dx term comes about as a result of the chain rule, which states that d(f
)/dx = df/dy * dy/dx. For instance, when we take the derivative of the first term with respect to x,d(y^4)/dx
we write this as:
d(y^4)/dy * dy/dx
d(y^4)/dy is 4y^3, so that first term is 4y^3 * dy/dx. When you're taking the derivative of a term that contains two functions, it's convenient to use the product rule, which states:
d(f(x)g(x))/dx = f'(x)g(x) + f(x)g'(x).
For example, taking the derivative of the third term, -3x^3 siny (the first function is -3x^3, the second is siny),
d(-3x^3 siny)/dx = d(-3x^3)/dx * siny - 3x^3 * d(siny)/dx (product rule)
= - 9x^2 siny - 3x^3 * d(siny)/dy * dy/dx (chain rule on the second term, since it contains y instead of x)
= - 9x^2 siny - 3x^3 cos
* dy/dxAnother question. I have a question to determine whether R is an equivalence relation or partial order. The problem is that it's neither. It's reflexive and symmetric but not transitive, which doesn't classify as equivalence or partial. What should I put down? Binary relation?
Yes, I am assuming the variance is the same. That's all the information given in the question, making me believe it's a matter of algebra to figure our this relationship.
The question is a bit weird... Why would you want wider confidence intervals?
In any case, the width of a confidence interval can be expressed as constant/sqrt(N), where N is the sample size. Hence, if you want to increase the width by 50% you need N* such that 1/sqrt(N*) is 50% bigger than 1/sqrt(N).
I think that would be correct if you used the value of r you found previously.
Yep, that was it. I was so scared to enter it as that was my last attempt
As you approach 5pi/2 from above, tan x tends to - infinity (if you were approaching 5pi/2 from the left rather than right, tan x would be tending to +infinity).
As you approach 5pi/2 from above, tan x tends to - infinity (if you were approaching 5pi/2 from the left rather than right, tan x would be tending to +infinity).
This whole + and - approach thing is ass backwards. You'd think the + would mean follow to the right of the graph, in which case it's heading towards infinity :/
Self-Loathing Narcissist
Banned
tan x = sin x/cos x
as x→(5π/2), tan x → tan(5π/2) = sin(5π/2)/cos(5π/2) = 1/0
as x approaches (5π/2), the limit blows up to either negative or positive infinity. For f(x) = tan x, there are vertical asymptotes present at multiples of π after π/2 (at π/2, 3π/2, 5π/2, etc.) Looking at the right-handed limit as x approaches 5π/2 from the right, the limit would approach negative infinity.
The question is a bit weird... Why would you want wider confidence intervals?
In any case, the width of a confidence interval can be expressed as constant/sqrt(N), where N is the sample size. Hence, if you want to increase the width by 50% you need N* such that 1/sqrt(N*) is 50% bigger than 1/sqrt(N).
I think it's more a conceptual question than anything, that's why the prof was asking. But thanks for your help. I'll play aroudn with it a little more. Solutions come up Weds, so I'll post the solution soon.
I always remember it by thinking of the + as meaning "slightly greater than this value", so x would have to be approaching from the right.
That helps!
Permanently A
Junior Member
(4+cosx)(2xcosx+2sinx) - (2xsinx)(-sinx)
By expanding I got:
8xcosx + 8sinx + 2xcos^2(x) + 2sinxcosx - 2xsin^2(x)
However the final answer is:
2sinxcosx + 8xcox + 8sinx + 2x
So what happened to the 2xcos^2(x) and - 2xsin^2(x) terms, and where did the 2x come from?
By expanding I got:
8xcosx + 8sinx + 2xcos^2(x) + 2sinxcosx - 2xsin^2(x)
However the final answer is:
2sinxcosx + 8xcox + 8sinx + 2x
So what happened to the 2xcos^2(x) and - 2xsin^2(x) terms, and where did the 2x come from?
the middle expression should have +2xsin^2(x).
look up trigonometric identities for your second question.
Permanently A
Junior Member
Got it, thanks!
I'm doing an interval of convergence problem and I was wondering if anyone would help me with the simplification.
The given series is
(2n + 1)!(x-5)^n
--------------------------
n!
So using the ratio test you get
(2(n+1)+1)!(x-5)^n+1 n!
------------------------------- x ----------------
(n+1)! (2n+1)!(x-5)^n
So far I have that going down to
(2n+3)!(x-5)
------------------
(n+1)(2n+1)!
How do I get rid of those last two factorials?
The given series is
(2n + 1)!(x-5)^n
--------------------------
n!
So using the ratio test you get
(2(n+1)+1)!(x-5)^n+1 n!
------------------------------- x ----------------
(n+1)! (2n+1)!(x-5)^n
So far I have that going down to
(2n+3)!(x-5)
------------------
(n+1)(2n+1)!
How do I get rid of those last two factorials?
So then that gives me (x-5) lim (2n+3)(2n+2)/n+1
That limit goes to infinity.
Then I have (x-5) * infinity.
What does that mean for the inteval of convergence?
For the examples we did in class, the limit part always converged to a finite number and then we got endpoints using the x part. I don't know what to do if the limit is infinity.
That limit goes to infinity.
Then I have (x-5) * infinity.
What does that mean for the inteval of convergence?
For the examples we did in class, the limit part always converged to a finite number and then we got endpoints using the x part. I don't know what to do if the limit is infinity.
If the limit is infinity, then the ratio test tells you that the series never converges. But in the case that x=5, the limit is zero, so the ratio test tells you that the series does still converge there, and only there.
As Therion said, that means that the series never converges (outside of x = 5).
However, you didn't need to use the ratio test for this. Why not? Remember that if a series converges, then the general term a_n must converge to 0. In your case, for any x !=, 5 it's easy to see that a_n = (2n+1)!/n! (x-5)^n does not converge to 0, which means that the series can't converge.
Just offering a different view!
Ok Thanks.
Next I'm trying to find the power series representation of the function 2x/(7+x^3)
I rewrote everything as
2x/7 (1/1- -x^3/7)
then
2x/7 series 0 to infinity (-1)^n(x^3/7)^n
then
series 0 to infinity (-1)^n (2x^3n+1)/7^n+1
Did I do that right?
edit:
Wolframalpha seems to be aligning with what I have.
However when I attempt to find the interval of convergence, all the of n-terms cancel and I'm only left with x's and constants. Does that mean the limit just goes away and I evaluate the absolute value like normal?
Next I'm trying to find the power series representation of the function 2x/(7+x^3)
I rewrote everything as
2x/7 (1/1- -x^3/7)
then
2x/7 series 0 to infinity (-1)^n(x^3/7)^n
then
series 0 to infinity (-1)^n (2x^3n+1)/7^n+1
Did I do that right?
edit:
Wolframalpha seems to be aligning with what I have.
However when I attempt to find the interval of convergence, all the of n-terms cancel and I'm only left with x's and constants. Does that mean the limit just goes away and I evaluate the absolute value like normal?
Biggest-Geek-Ever
Member
Yes. If all n's cancel then you only need to find the interval on which the absolute value of your resultant is < 1. Don't forget to check the endpoints!
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