Support NeoGAF

[youngwerther]

My teacher has a fun habit of giving us homework over things we haven't yet covered in class.

I'm trying to do Taylor polynomial approximations teaching myself with internet videos and as you can imagine it's not going well.

I'll just write out everything, please tell me what I'm doing wrong.

Question -

Find the degree 3 Taylor polynomial approximation to the function f(x) = 2ln(sec(x)) at a = 0.

From what I gather that means I need the first 3 derivatives.

So

f(x) = 2ln(sec(x))
f'(x) = 2tan(x)
f''(x) = 2sec^2(x)
f'''(x) = 4tan(x)sec^2(x)

and then I need the values of these at a = 0

So

f(0) = 0
f'(0) = 0
f''(0) = 2
f'''(0) = 0

Now, the formula I should be using is I guess this

f(a) + (f'(a)/1!)(x-a)^1 + (f''(a)/2!)(x-a)^2 + (f'''(a)/3!)(x-a)^3

When I plug everything in -

0 + (0/1)(x) + (2/2)(x)^2 + (0/6)(x)^3

So as far as I can tell, the zeros kill the 1st and 3rd degree terms, and I'm left with 2x^2.

That isn't correct. Anyone know what's wrong here?

 

[MikeDip]

My teacher has a fun habit of giving us homework over things we haven't yet covered in class.

I'm trying to do Taylor polynomial approximations teaching myself with internet videos and as you can imagine it's not going well.

I'll just write out everything, please tell me what I'm doing wrong.

Question -

Find the degree 3 Taylor polynomial approximation to the function f(x) = 2ln(sec(x)) at a = 0.

From what I gather that means I need the first 3 derivatives.

So

f(x) = 2ln(sec(x))
f'(x) = 2tan(x)
f''(x) = 2sec^2(x)
f'''(x) = 4tan(x)sec^2(x)

and then I need the values of these at a = 0

So

f(0) = 0
f'(0) = 0
f''(0) = 2
f'''(0) = 0

Now, the formula I should be using is I guess this

f(a) + (f'(a)/1!)(x-a)^1 + (f''(a)/2!)(x-a)^2 + (f'''(a)/3!)(x-a)^3

When I plug everything in -

0 + (0/1)(x) + (2/2)(x)^2 + (0/6)(x)^3

So as far as I can tell, the zeros kill the 1st and 3rd degree terms, and I'm left with 2x^2.

That isn't correct. Anyone know what's wrong here?

You're left with x^2, not 2x^2. Either way, it all looks fine, but perhaps the answer you have wants more terms? What is the answer you should have, since that is all correct.

 

[youngwerther]

holy SHIT I'm an idiot.

Wow. I need to go to bed. Thanks dude.

 

[MikeDip]

holy SHIT I'm an idiot.

Wow. I need to go to bed. Thanks dude.

Haha, I know the feeling dude, don't worry about it. Doing math when tired leads to all sorts of mistakes. You should be happy you got all that stuff right, especially since it wasn't covered in class yet.

 

[Exuro]

Alright I've got a pretty simple problem(but may be hard to explan with text). I have two functions that need to be convolved graphically. The thing is I haven't convolved a function graphically that is partially in the negative x axis(goes from -1 to 1). One function is only in the positive x axis which when I flip and move left of the other function and set the right side to t, but if I were to flip a function that is partially in the negative axis, do I make t where it was zero or the right most part?I tried doing the latter and when I tried flipping with one function and then the other I ended up getting different equations. I'll try to make a picture as this is a little wordy.

Edit:

Okay so if I flipped h(t), it would flip horizontally and be set outside to the left of g(t). The rightmost end of h(t) would be set as t while the left would be t-1. If I were to flip g(t), Would the rightmost be t or would it be where 0 is so the rightmost would actually be t+1?

 

[MikeDip]

Alright I've got a pretty simple problem(but may be hard to explan with text). I have two functions that need to be convolved graphically. The thing is I haven't convolved a function graphically that is partially in the negative x axis(goes from -1 to 1). One function is only in the positive x axis which when I flip and move left of the other function and set the right side to t, but if I were to flip a function that is partially in the negative axis, do I make t where it was zero or the right most part?I tried doing the latter and when I tried flipping with one function and then the other I ended up getting different equations. I'll try to make a picture as this is a little wordy.

Edit:


Okay so if I flipped h(t), it would flip horizontally and be set outside to the left of g(t). The rightmost end of h(t) would be set as t while the left would be t-1. If I were to flip g(t), Would the rightmost be t or would it be where 0 is so the rightmost would actually be t+1?

Are these two specific questions or one question? Because if you're taking h(x) as the one you are going to flip and slide along g(x) to get their multiplication for the integral then you don't have to flip g(x), right? You just have to change your boundaries of integration when solving.
Where are you flipping something that is partially negative in this question?

 

[Exuro]

Are these two specific questions or one question? Because if you're taking h(x) as the one you are going to flip and slide along g(x) to get their multiplication for the integral then you don't have to flip g(x), right? You just have to change your boundaries of integration when solving.
Where are you flipping something that is partially negative in this question?

The question asks to flip one and do the computations and then flip the other and do the computations and then compare that the convolved functions are the same. What I mean by negative is all of the examples I've done with functions they've always started at t=0 while x(t) starts at t=-1. I was wondering if when I flip x(t)(turning it into x(t-tau) and move it along h(t) from the left where I set t to on x(t-tau), be it on the right most side of the function or where it was at zero and the rightmost would become t+1.

 

[MikeDip]

The question asks to flip one and do the computations and then flip the other and do the computations and then compare that the convolved functions are the same. What I mean by negative is all of the examples I've done with functions they've always started at t=0 while x(t) starts at t=-1. I was wondering if when I flip x(t)(turning it into x(t-tau) and move it along h(t) from the left where I set t to on x(t-tau), be it on the right most side of the function or where it was at zero and the rightmost would become t+1.

If I recally correctly, the rightmost would be t+1.

 

[Exuro]

If I recally correctly, the rightmost would be t+1.

Okay, so t is always set to where the function was originally 0? If so I think I got it all correct now.

 

[MikeDip]

Okay, so t is always set to where the function was originally 0? If so I think I got it all correct now.

Yeah, did you get the same answer when you compared both answers now?

 

[Exuro]

Yeah, did you get the same answer when you compared both answers now?

Yeah I got the same set of equations so each graph looks the same. Before I put up this question I set the rightmost side to t and then when I went to flip the other function I realized my equations were turning out very different, so now that I know to set t to where it was originally zero everything came out right.

 

[MikeDip]

Yeah I got the same set of equations so each graph looks the same. Before I put up this question I set the rightmost side to t and then when I went to flip the other function I realized my equations were turning out very different, so now that I know to set t to where it was originally zero everything came out right.

Awesome, happy it worked out for you.

 

[RyanDG]

So, I have a calculus test tomorrow and I am running into one snag on the review sheet that doesn't seem to want to behave right for me.

I'm asked to integrate the following expression over the interval (1, 9):

square root (7/z) dz.

I'm assuming its going to be a substitution problem of some kind, but for some odd reason, I'm not coming up with a substitution technique that is working right.

I know its probably extremely simple, but after being able to work through the last 35 problems without a hitch, I'm sort of at loss as to why I just can't see the solution to this one.

Help?

 

[MikeDip]

So, I have a calculus test tomorrow and I am running into one snag on the review sheet that doesn't seem to want to behave right for me.

I'm asked to integrate the following expression over the interval (1, 9):

square root (7/z) dz.

I'm assuming its going to be a substitution problem of some kind, but for some odd reason, I'm not coming up with a substitution technique that is working right.

I know its probably extremely simple, but after being able to work through the last 35 problems without a hitch, I'm sort of at loss as to why I just can't see the solution to this one.

Help?

Doing u = sqrt(z) works fine.

EDIT: more info...

This gives du=(1/2)z^(-1/2)dz. Sub into integral, and you get
sqrt(7)(2)z^(1/2)(1/u)du But now can sub the z^(1/2) for u again. Whole thing simplifies to
2sqrt(7)du. Now just integrate and sub back in for z like usual.
Gives 2*sqrt7*sqrtz, now just put in your endpoints and get the number you need.

 

[opticalmace]

So, I have a calculus test tomorrow and I am running into one snag on the review sheet that doesn't seem to want to behave right for me.

I'm asked to integrate the following expression over the interval (1, 9):

square root (7/z) dz.

I'm assuming its going to be a substitution problem of some kind, but for some odd reason, I'm not coming up with a substitution technique that is working right.

I know its probably extremely simple, but after being able to work through the last 35 problems without a hitch, I'm sort of at loss as to why I just can't see the solution to this one.

Help?

rt(7/z) = rt(7) / rt(z)

i.e. = constant * z^(-1/2)

 

[Lion Heart]

Alright GAF.

Somehow 6[(2^1/2)x]^5(3^1/2) becomes 24√6x^5

How?

Heres a Paint version

 

[MikeDip]

Alright GAF.

Somehow 6[(2^1/2)x]^5(3^1/2) becomes 24√6x^5

How?

Heres a Paint version

2^(5/2) equals 4*√2. (since 2^5/2 = 2^(4/2)*2^(1/2)=(2^2)*√2= 4√2
4√2 * √3 = 4*√(2*3) = 4*√6.
So when all multiplied together you get:
4*√6 * 6 * x^5 = 24√6x^5

 

[Lion Heart]

Oh, interesting. So when you have √2^5 does that =√32 = √16√2=4√2? I guess thats another way to do it?

 

[MikeDip]

Oh, interesting. So when you have √2^5 does that =√32 = √16√2=4√2? I guess thats another way to do it?

Yeah, that's fine.
Just remember that √2 is the same as 2^(1/2). So (√2)^5 = 2^(5/2)= 2^(5*(1/2)) = (2^5)^(1/2) = √(2^5)

 

[RyanDG]

rt(7/z) = rt(7) / rt(z)

i.e. = constant * z^(-1/2)

Doing u = sqrt(z) works fine.

EDIT: more info...

This gives du=(1/2)z^(-1/2)dz. Sub into integral, and you get
sqrt(7)(2)z^(1/2)(1/u)du But now can sub the z^(1/2) for u again. Whole thing simplifies to
2sqrt(7)du. Now just integrate and sub back in for z like usual.
Gives 2*sqrt7*sqrtz, now just put in your endpoints and get the number you need.

Thanks guys, I knew my mind was just being dumb there.

Didn't even realize the substitution wasn't even necessary if you treat it like above.

 

[Ultryx]

The polynomial function f has exactly one positive zero. Approximate the zero correct to two decimal places. Round to two decimal places.

f(x)= 2x^4 - 20x^3 - 9x^2 - 10x - 5

I'm taking Algebra 2 online and the program uses MyMathLabs, which is normally super helpful, but I cannot understand how to solve this problem. I'm at a loss for the example of what the Bounds on Zeros Theorem is, too.

 

[cpp_is_king]

If you can find a value of x where f(x) is negative, and another where f(x) is positive, your zero must be somewhere in between.

What is f(0)? f(1)? f(2)? Once the sign changes, try to tighten the bound further

 

[Exuro]

Question for those that have dealt with fourier series. I'm taking a function that is discontinuous in two places and am using fourier series partial sums to create a new graph for it in the 3rd harmonic. I'm pretty sure I've entered all the data in correctly, as the maximum and minimum are roughly the same as on the non-fourier graph, but for some reason the endpoints are very off. My graph goes from x=0->36 and y=o->20. The end points are hitting 13 when the original is 10. Is this a thing that fourier series does or do I need to go over my data and double check that it's right? This is the first fourier series I've done so I don't know a whole lot other than plugging into the equation.

 

[Leezard]

Question for those that have dealt with fourier series. I'm taking a function that is discontinuous in two places and am using fourier series partial sums to create a new graph for it in the 3rd harmonic. I'm pretty sure I've entered all the data in correctly, as the maximum and minimum are roughly the same as on the non-fourier graph, but for some reason the endpoints are very off. My graph goes from x=0->36 and y=o->20. The end points are hitting 13 when the original is 10. Is this a thing that fourier series does or do I need to go over my data and double check that it's right? This is the first fourier series I've done so I don't know a whole lot other than plugging into the equation.

How many terms do you have in the Fourier series? For a closer fit, you might need more terms.

 

[Exuro]

How many terms do you have in the Fourier series? For a closer fit, you might need more terms.

Well I did an increment of 1 with a period of 36 so theres 36 terms(37 including t=0) for Ao An Bn. It's also set to the 3rd harmonic. Ive checked my data and it all looks correct so Im not sure why the endpoints would be of or why at t=0 it has a negative slope. Here's a sketch of the waveform and my fourier series. Sorry for the terrible proportions. Am I on the right track here? I honestly have no idea as this is the only assignment I've done using fourier.


I just noticed that the max/min arent at the same t values either... hmm...

 

[Leezard]

Well I did an increment of 1 with a period of 36 so theres 36 terms(37 including t=0) for Ao An Bn. It's also set to the 3rd harmonic. Ive checked my data and it all looks correct so Im not sure why the endpoints would be of or why at t=0 it has a negative slope. Here's a sketch of the waveform and my fourier series. Sorry for the terrible proportions. Am I on the right track here? I honestly have no idea as this is the only assignment I've done using fourier.


I just noticed that the max/min arent at the same t values either... hmm...

I see, I did not mean the data points when I said terms, I meant the order of the Fourier series. When you say it's set to the third harmonic, do you mean that the Fourier series looks like a0/2 + a1cos(t) + b1sin(t) + a2cos(2t) + b2sin(2t) +a3cos(3t) + b3sin(3t)? Because if so, then that's most likely the reason you won't get a closer fit. If you have a high enough order on the Fourier series (typically around half the amount of data points) you should be able to get a near perfect fit.

Since it looks like you do this in Matlab, you can do a simple Fourier series fit with these files, just to check that you've indeed done everything right.
https://www.mathworks.com/matlabcentral/fileexchange/31013-simple-real-fourier-series-approximation

 

[Exuro]

I see, I did not mean the data points when I said terms, I meant the order of the Fourier series. When you say it's set to the third harmonic, do you mean that the Fourier series looks like a0/2 + a1cos(t) + b1sin(t) + a2cos(2t) + b2sin(2t) +a3cos(3t) + b3sin(3t)? Because if so, then that's most likely the reason you won't get a closer fit. If you have a high enough order on the Fourier series (typically around half the amount of data points) you should be able to get a near perfect fit.

Since it looks like you do this in Matlab, you can do a simple Fourier series fit with these files, just to check that you've indeed done everything right.
https://www.mathworks.com/matlabcentral/fileexchange/31013-simple-real-fourier-series-approximation

Very cool. Ill be sure to mess around with that after i leave campus. I went and set up the data for the 20th term/harmonic so hopefully it'll resemble better.

 

[Lion Heart]

y=sin40 has a period of 90deg. How is that exactly?

 

[big ander]

y=sin40 has a period of 90deg. How is that exactly?

I'm assuming "0" is theta or x (if it's really sine of forty then y is just constant).
Then in general you know that for sin(ax) the period is 2pi/a. Here a=4, so 2pi/4=pi/2 in radians which is 90 degrees.
If it helps to plot it out, just take what you know about the general version sin(x) and apply it to what you have. You know that something of the form sin(x) is going to have one period from x=0 to x=2pi. Here you have x=4theta. so starting at 4theta=0 (thus theta=0), a period is complete at 4theta=2pi => theta=2pi/4=pi/2.

 

[Timedog]

Couldn't they just take all the trig stuff out of Calculus and find some other more intuitive way of expressing those things based off glyphs found in the Roswell crash or something? I fucking hate trig and I fucking hate parametric bullshit and I want it out of my Calculus series because I'm not a fan of having to actually think.

 

[Lion Heart]

Ohhh jesus, I read it as sin40 and was confused as hell. It is 4 theta so now it makes sense. Honestly I have no idea how your method works but for me 4 is the K value and the period is just 360/k. a would be in front of sin. Thanks though.

 

[big ander]

Ohhh jesus, I read it as sin40 and was confused as hell. It is 4 theta so now it makes sense. Honestly I have no idea how your method works but for me 4 is the K value and the period is just 360/k. a would be in front of sin. Thanks though.

yeah I mean 2pi radians = 360 degrees. If your trig course hasn't covered conversion between degrees and radians yet I would hope it's coming up soon.

 

[Tater Tot]

hey can someone help me out please ?

Evaluate the intregal using an appropriate substitution

(dx)/(squareroot(1-36x^(2)))

 

[krameriffic]

hey can someone help me out please ?

Evaluate the intregal using an appropriate substitution

(dx)/(squareroot(1-36x^(2)))

Has the form of a trigonometric substitution integral. You write it as:

dx/6(root((1/6)^2-x^2))

By factoring out the 36 from the denominator and taking the square root to pull it out of the radical on the bottom (hence the factor of 1/6). At this point it's in the form of 1/root(a^2-x^2), where you would substitute x = asin, dx = acosdy (where a = 1/6). Substituting that back in and using some trig identities yields:

(1/6)(dy)acos/root((acos)^2) = (1/6)(dy)

Integration yields y/6 + C, and inverting the substitution yields:

(1/6)arcsin(6x) + C

As the final answer. I think.

 

[youngwerther]

I took a test and got a True/False question wrong and I can't figure out why.

It says -

"For functions f(x) and g(x), limx->a (f(x) + g(x)) = lim x -> a f(x) + lim x ->a g(x).

I put True and it was marked wrong.

Why is it false? To me it seems to be simply saying that the limit of the sum is the sum of the limits.

 

[Leezard]

I took a test and got a True/False question wrong and I can't figure out why.

It says -

"For functions f(x) and g(x), limx->a (f(x) + g(x)) = lim x -> a f(x) + lim x ->a g(x).

I put True and it was marked wrong.

Why is it false? To me it seems to be simply saying that the limit of the sum is the sum of the limits.

Typically stuff like this only blows up in the case of limits not existing.
Let f = 1/x, g = -1/x, a = 0. The limit of the sum exists, but the individual limits diverge.

 

[Tater Tot]

Has the form of a trigonometric substitution integral. You write it as:

dx/6(root((1/6)^2-x^2))

By factoring out the 36 from the denominator and taking the square root to pull it out of the radical on the bottom (hence the factor of 1/6). At this point it's in the form of 1/root(a^2-x^2), where you would substitute x = asin, dx = acosdy (where a = 1/6). Substituting that back in and using some trig identities yields:

(1/6)(dy)acos/root((acos)^2) = (1/6)(dy)

Integration yields y/6 + C, and inverting the substitution yields:

(1/6)arcsin(6x) + C

As the final answer. I think.

Yeah, I actually got the same answer but somehow it was wrong. Thank you though.

 

[Tater Tot]

Double post sorry:

Definite Integral

Upperlimit (4) / lower limit (0) Original problem = (2x+1)^(3)

for the antiderivative I got (1/8)(2x+1)^(4)

after plugging in the limits I got (6561/8)-(2401/8)=520

Some how it says it is wrong. Did I do something wrong?

Edit: Nevermind I solved it

 

[Exuro]

I see, I did not mean the data points when I said terms, I meant the order of the Fourier series. When you say it's set to the third harmonic, do you mean that the Fourier series looks like a0/2 + a1cos(t) + b1sin(t) + a2cos(2t) + b2sin(2t) +a3cos(3t) + b3sin(3t)? Because if so, then that's most likely the reason you won't get a closer fit. If you have a high enough order on the Fourier series (typically around half the amount of data points) you should be able to get a near perfect fit.

Since it looks like you do this in Matlab, you can do a simple Fourier series fit with these files, just to check that you've indeed done everything right.
https://www.mathworks.com/matlabcentral/fileexchange/31013-simple-real-fourier-series-approximation

Hey wanted to say thanks for the help. I have one last question that's only somewhat related to the problem. I did my fourier up to the 20th term and I'm wondering if there's a way to plot the function smoothly. What I mean by smoothly is that when I use plot(t, func) it only takes data from 0-36 and then runs "connect the dots" instead of looking like small waveforms.

This is what it looks like right now. I saw a friends waveform last week and it looked nice and wavy and we used the same data and method to get the equation.

EDIT: nevermind didn't realize how easy it was to change t.

 

[A good poster]

What's F'(x) if F(x) = uplimit.= cosx, lowlimit =sinx int. sqroot t^2+1?

I have no idea how to write down integrals in a coherent manner on a keyboard.

 

[Tater Tot]

Solve the intial value problem

(dy)/(dt)=1/t y(-1)=4

please help

 

[Leezard]

Solve the intial value problem

(dy)/(dt)=1/t y(-1)=4

please help

If the problem is correctly written, then it is unsolvable since ln(-1) is undefined. Edit: oh, right, you could use the absolute value, but the problem is still badly defined.

 

[0xCA2]

Hey GAF. I got -12.22 on this question, is that right?

First I looked for the angle between the two vectors. I subtracted 12 from 360 to get 333 degrees ( the positive version of the negative angle there ) and I added 180 to 117 to get 297 ( the full angle of the reference angle ). I subtracted 297 from 333 to get 136 degrees. I then multiplied the cos of that by 17 ( the magnitude of u ) and got -12.22. Is this right?

 

[luoapp]

Hey GAF. I got -12.22 on this question, is that right?



First I looked for the angle between the two vectors. I subtracted 12 from 360 to get 333 degrees ( the positive version of the negative angle there ) and I added 180 to 119 to get 297 ( the full angle of the reference angle ). I subtracted 297 from 333 to get 136 degrees. I then multiplied the cos of that by 17 ( the magnitude of u ) and got -12.22. Is this right?

It can't be negative.

 

[0xCA2]

Hey GAF. I got -12.22 on this question, is that right?



First I looked for the angle between the two vectors. I subtracted 12 from 360 to get 333 degrees ( the positive version of the negative angle there ) and I added 180 to 117 to get 297 ( the full angle of the reference angle ). I subtracted 297 from 333 to get 136 degrees. I then multiplied the cos of that by 17 ( the magnitude of u ) and got -12.22. Is this right?

Yeah, this is been busting my brain for a bit and I just realized that I accidentally thought I got 136 degrees when I got 36 degrees. Using that, I end up with 13.75, I think that should be right.

 

[Exuro]

Okay I'm confused with this second order system in linear algebra.

I'm given

I was able to find the four eigenvalues and vectors that I needed(They are correct). My question is why are the second two terms in the general solution equation negative?

I'm looking at examples in the book and none of them have negative terms. I'm completely lost as the only issue is the negative signs.

 

[Partial Gamification]

Okay I'm confused with this second order system in linear algebra.

I'm given

I was able to find the four eigenvalues and vectors that I needed(They are correct). My question is why are the second two terms in the general solution equation negative?

I'm looking at examples in the book and none of them have negative terms. I'm completely lost as the only issue is the negative signs.

My understanding (could be wrong) is that the constants are typically written as positive terms. The step is usually skipped - the negative is dropped, by subbing the negative term for another constant.
-1*C_3 = C_k
-1*C_4 = C_j
The reasoning is that they are constant coefficients, choosing a negative term would just make the sign reverse. This is to say that from the general solution, one will choose the coefficients and ultimately determine their signage through choosing. I want to call it a sort of standard reduction but I'm not certain.

edit: this might help:
constant*constant = constant

 

[MikeDip]

What's F'(x) if F(x) = uplimit.= cosx, lowlimit =sinx int. sqroot t^2+1?

I have no idea how to write down integrals in a coherent manner on a keyboard.

I missed this question before, sorry. Do you still need help with this?

 

[decaf]

Probably not difficult for any of you lot, but I'm trying to figure out the largest box I can send using the above rules.

Any ideas for vaguely "normal" dimensions?

 

[Leezard]

Largest in terms of volume, I guess. Make a function of volume depending on length, width and height, and differentiate after simplifying to a single variable, with the help of the constraints.