The math help thread

I know this is probably a really simple question, but no matter what I do I get the wrong answer! =[

So if anybody can help, that'd be great.

A laboratory measurement system used to measure blood potassium levels has a standard deviation of *sigma* = 0.2 mEq/dl in repeated testing. Repeated readings are normally distributed with mean equal to the true potassium level in the blood sample. How many times should the blood samples be tested for one person to get a margin of error no larger than *plus/minus* 0.15 mEq/dl with 95% confidence?

I know this is probably a really simple question, but no matter what I do I get the wrong answer! =[

So if anybody can help, that'd be great.

A laboratory measurement system used to measure blood potassium levels has a standard deviation of *sigma* = 0.2 mEq/dl in repeated testing. Repeated readings are normally distributed with mean equal to the true potassium level in the blood sample. How many times should the blood samples be tested for one person to get a margin of error no larger than *plus/minus* 0.15 mEq/dl with 95% confidence?

I have to look in my stats book but I think for confidence interval it's

sample mean/average "+ or -" sigma/SQRT(n) * Z(alpha)

Sample mean/average is "u" or "X bar"

Sigma in this problem is 0.2

SQRT(n) is square root of sample

Z(alpha) relates to 95%, in this case it's about 1.96 if i'm not mistaken

So just solve for 'n' i.e.

sigma/SQRT(n) * 1.96 = 0.15

----

I might have some things wrong because I don't wanna dig up my notes, but it's just a formula somewhere.

I have to look in my stats book but I think for confidence interval it's

sample mean/average "+ or -" sigma/SQRT(n) * Z(alpha)

Sample mean/average is "u" or "X bar"

Sigma in this problem is 0.2

SQRT(n) is square root of sample

Z(alpha) relates to 95%, in this case it's about 1.96 if i'm not mistaken

So just solve for 'n' i.e.

sigma/SQRT(n) * 1.96 = 0.15

----

I might have some things wrong because I don't wanna dig up my notes, but it's just a formula somewhere.

Thanks, sir. I know your answer is right because I went to the office hours earlier today and had it explained to me more clearly

I got the 95% confidence mistaken since I directly took 95% from the normal table when I should have taken half of 100%-95%, which is 1.96. *Understands the material now *

try being less of a f'aggot

Hi im in calculus AB in my school and im struggling with inverse functions >_>. I can do the basics but when the problem is like y = 5x^3+5x and im told to find the inverse i really freeze up. I take the derivative but i forget what to do afterwards :>

Hi im in calculus AB in my school and im struggling with inverse functions >_>. I can do the basics but when the problem is like y = 5x^3+5x and im told to find the inverse i really freeze up. I take the derivative but i forget what to do afterwards :>

Here's a basic equation for inverses:



Yes, you do find the derivative, as you'll need it later. However, you need to also find the inverse. This can be achieved in 2 ways

1) Switch x and y, and solve for f^(-1)(x).

2) If given a condition, like "Given that g'(x) = f^(-1)(x) and g'(1) = 2, find the inverse of f(x) = 5x^6 + 10x^4 + 2x^3", that's another method you can do it (this is mostly found in AP exams). This is most common when you cannot normally inverse an function (and a condition is given to you), or in place of a defined function, it gives you discreet conditions like "Given f'(5) = 10 and g'(3) = 5 and f^(-1)(X) = g'(x)"....etc


Got any specific examples you want us to help you on?

Hi, I'm totally sucking at math right now, particularly differential equations. Right now, I'm in the section of the book talking about solving nonhomogeneous equations suing the method of undetermined coefficients.
A specific problem I'm having trouble with is y"-3y'+2y = e^x*sin(x). Anyone care to explain the process to me?

The answer in the back of the book is y = c1*e^x + c2*e^(2x) + 1/2*e^x*(cos(x)-sin(x)) where c1 and c2 are arbitrary coefficients.

Hi, I'm totally sucking at math right now, particularly differential equations. Right now, I'm in the section of the book talking about solving nonhomogeneous equations suing the method of undetermined coefficients.
A specific problem I'm having trouble with is y"-3y'+2y = e^x*sin(x). Anyone care to explain the process to me?

The answer in the back of the book is y = c1*e^x + c2*e^(2x) + 1/2*e^x*(cos(x)-sin(x)) where c1 and c2 are arbitrary coefficients.

Haven't done this in a while, but I'll give it a shot.


y"-3y'+2y = e^x*sin(x)

First solve as if it was homogeneous

so we have y''-3y'+2y=0
Using the quadratic formula, on m^2 -3m +2

3 +/- sqrt(9-4*1*2) /2
So m= (3 +sqrt1)/2 = 2
or m = (3- sqrt 1)/2 = 1

With the 2 roots, we now have a general solution with 2 arbitary constants, c1 and c2 being

y= c1*e^(x) + c2*e^(2x)

Then we need to find a complementary solution for the full equation ( I admit, I had to check a list of complementary solutions to know which one to use here)
If we look at the right hand side of the original equation, e^x *sin(x), we should try a particular solution of the form Ae^x*cosx + Be^x*sinx, (where A and B are constants, but they CAN be 0)
So, we need to differentiate this twice using hte product rule (ugh, horrible I know)
letting y= Ae^x*cosx +Be^x*sinx


y' = Ae^x*cos(x)-Ae^x*sin(x) + Be^x*sin(x) + Be^x*cos(x)

y'' = Ae^x*cos(x)-Ae^x*sin(x) -Ae^x*sin(x)-Ae^x*cos(x) + Be^x*sin(x)+Be^x*cos(x) +Be^x*cos(x)-Be^x*sin(x)
which cancels nicely to
y''= -2Ae^x*sin(x) +2Be^x*cos(x)

Substituting these values for y, y' and y'' back into the original equation y"-3y'+2y = e^x*sin(x)

Gives
-2Ae^x*sin(x) +2Be^x*cos(x) -3[Ae^x*cos(x)-Ae^x*sin(x) + Be^x*sin(x) + Be^x*cos(x)] + 2[Ae^x*cosx +Be^x*sinx] = e^x*sin(x)

Then we can cancel the e^x which is in every term to get

-2A*sin(x) +2Bcos(x) -3Acos(x)+3Asin(x) -3Bsin(x) -3Bcos(x) + 2Acosx +2Bsinx = sin(x)

Then we can gather all the sin and cos terms on both sides seperatly to get 2 equations

-2Asin(x) +3Asin(x) -3Bsin(x) +2Bsinx = sin(x)
=> Asin(x)-Bsin(x)= sin(x)
=> A-B = 1

And
2Bcos(x) -3Acos(x) -3Bcos(x) + 2Acosx = 0
(Dividing both sides by cos(x)]
=>2B-3A-3B+2A = 0
=> -A-B = 0

Solving these simultaneously gives
-2 B = 1 => B= -1/2
A= 1/2

Throw these back into our complementary solution, and bobs your uncle
Ae^x*cosx +Be^x*sinx becomes
1/2e^x*cos(x)-1/2 e^x *sin(x)
= 1/2 e^x (cos(x)-sin(x))

Then combining this with our general solution, this becomes the complete solution to the equation.

y= c1*e^(x) + c2*e^(2x) +1/2 e^x (cos(x)-sin(x)).

If I made any mistakes in my working, I blame it on doing this all in my head instead of using paper like a sane person

Hi, I'm totally sucking at math right now, particularly differential equations. Right now, I'm in the section of the book talking about solving nonhomogeneous equations suing the method of undetermined coefficients.
A specific problem I'm having trouble with is y"-3y'+2y = e^x*sin(x). Anyone care to explain the process to me?

The answer in the back of the book is y = c1*e^x + c2*e^(2x) + 1/2*e^x*(cos(x)-sin(x)) where c1 and c2 are arbitrary coefficients.

Have to remember what undetermined coefficient process was. It should be straightforward if you follow how a basic example works.

too bad I forgot the process f3

Oh, thanks Lozmaster! I just have trouble finding the particular solution of these equations. I never get it right -_-.

Oh, thanks Lozmaster! I just have trouble finding the particular solution of these equations. I never get it right -_-.

Like most maths, practice makes perfect, keep at it and you'll get it eventually. It took me too long trying to memorize most of the process, and I only actually managed it after that modules exam ;-;

Studying for calc, I stumbled upon these questions, could anyone show me how to solve these:

integrate: sin^3(x)cos^5(x)dx
integrate: sin^4(x)cos^5(x)dx

u = sin x
du = cos x dx

sin^3(x)cos^5(x)dx = u^3*(1-u^2)^2 du = u^3 - 2u^5 + u^7 du -> 1/4 u^4 - 2/6 u^6 + 1/8 u^8 + C
sin^4(x)cos^5(x)dx = u^4*(1-u^2)^2 du = u^4 - 2u^6 + u^8 du -> 1/5 u^5 - 2/7 u^7 + 1/9 u^9 + C

etc

To further append on, I'd like to take the general case for


Usually, one put this into three different cases: When m is odd, when m is even and n odd, and when m and n are even.

For the first case, m is odd. We can thus represent m = 2k + 1. Remembering the identity

we obtain the following:


Then we just combine the single sin with dx, and set sin x dx = -d(cos x)

For the second case, use the exact same way. Just swap cos and sin, and set cos x dx = d * sin x

For the third case, we'll should use another identity:


Then, just substitute this in, and there you go.

Noah

Who wants to explain why the derivatve of lnx is 1/x, or why the indefinite integral of 1/x is lnx.

I'm only in Algebra2, I'm just curious. :3

Kinda off-topic, but suddenly I'm really into Math nowadays. I mean, it's more addictive than gaming! Who knew studying could be so fun? Currently on Differenciation, just finished Trigo. Everyone's screaming in class but I realize that both topics are really enjoyable after you grasp the basic concept. People are mixing tediousness and difficulty.

Who wants to explain why the derivatve of lnx is 1/x, or why the indefinite integral of 1/x is lnx.

I'm only in Algebra2, I'm just curious. :3

http://en.wikipedia.org/wiki/Natural_log...lor_series

Iunno what you can and cannot read in Algebra2, but I think that might what you're after. Typical Wiki-Jargon, though. :/


Edit: ^Mildly related.
Found it in English.
http://www.ltcconline.net/greenl/courses...vative.htm

Intuitively:
y = e^x is an exponential, so dy/dx is proportional to y. (This is a property of exponentials.) Consider dx/dy: it's the reciprocal of dy/dx so it's proportional to 1/y.

Now switch the names of the variables to find the inverse. x = e^y, so y = ln x. And dy/dx is proportional to 1/x.

(If you start with the premise that d(e^x)/dx = e^x, not just proportional (which is true though I haven't shown it), then you get that d(ln x)/dx = 1/x. But hopefully the above at least gives you an idea.)

Under what conditions is a^b > b^a, and vise versa? Using 5 and 7 for a and b, 5^7 vs 7^5. 5^7 is larger. What do I have to change b to so that b^5 is larger? Et cetera.
Tried doing some algebra on this and got lost. Fast.

a^b > b^a
ln(a^b) > ln(b^a) -- assuming both are positive
b ln a > a ln b
(ln a)/a > (ln b)/b
Define f(x) = (ln x)/x
f(a) > f(b)

a^b > b^a if f(a) > f(b).

f(5) = 0.322
f(7) = 0.278

Plot of the function:


For x > e, f(x) is decreasing, so a^b > b^a if b > a > e.

Anything else in particular?