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 Calculus Calculus Math Forum

September 19th, 2016, 01:16 PM   #11
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 Originally Posted by JeffM1 I wanted to get rid of the fractions. Fractions are a pain. So I multiplied both sides of the equation by $C^2.$ $U = \dfrac{D}{C} + \dfrac{S}{C} * \dfrac{S}{C} * U = \dfrac{D}{C} + \dfrac{S^2U}{C^2} \implies$ $C^2 * U = C^2 \left ( \dfrac{D}{C} + \dfrac{S^2U}{C^2} \right ) \implies$ $C^2U = \dfrac{\cancel C * C * D}{\cancel C} + \dfrac{\cancel {C^2} S^2U}{\cancel {C^2}} \implies C^2U = CD + S^2U.$
i seem to forget "multiply on both sides" to ease the pain level of the equation ^^ September 19th, 2016, 01:42 PM   #12
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 Originally Posted by JeffM1 I wanted to get rid of the fractions. Fractions are a pain. So I multiplied both sides of the equation by $C^2.$ $U = \dfrac{D}{C} + \dfrac{S}{C} * \dfrac{S}{C} * U = \dfrac{D}{C} + \dfrac{S^2U}{C^2} \implies$ $C^2 * U = C^2 \left ( \dfrac{D}{C} + \dfrac{S^2U}{C^2} \right ) \implies$ $C^2U = \dfrac{\cancel C * C * D}{\cancel C} + \dfrac{\cancel {C^2} S^2U}{\cancel {C^2}} \implies C^2U = CD + S^2U.$
how did you arrive at that last expression ?

$\displaystyle U= \frac {D}{C- \frac{S^{2}}{C}}$

Last edited by Ku5htr1m; September 19th, 2016 at 01:53 PM. September 20th, 2016, 05:43 AM #13 Member   Joined: Sep 2013 Posts: 83 Thanks: 0 One more thing, i picked up what i believe is an error on that same slide: From this: $\displaystyle \beta *l+\sigma *\left \{ \frac{\alpha -p_{x}-\sigma *l}{\beta } \right \}=\alpha -p_{l}$ To this: $\displaystyle l=\frac{\beta -\sigma }{\beta ^2-\sigma ^2}*\alpha -\frac{\beta }{\beta ^2-\sigma ^2}*p_{l}+\frac{\sigma }{\beta ^2-\sigma ^2}*p_{x}$ i believe it should be : $\displaystyle l=\frac{\beta -\sigma }{\beta -\sigma ^2}*\alpha -\frac{\beta }{\beta -\sigma ^2}*p_{l}+\frac{\sigma }{\beta -\sigma ^2}*p_{x}$ September 20th, 2016, 07:49 PM #14 Senior Member   Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry I wonder how you can use LaTeX perfectly fine to write fractions but not for the multiplication symbol ($\displaystyle \times$). September 21st, 2016, 07:06 AM   #15
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 Originally Posted by Monox D. I-Fly I wonder how you can use LaTeX perfectly fine to write fractions but not for the multiplication symbol ($\displaystyle \times$).
?? September 21st, 2016, 07:20 AM   #16
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 Originally Posted by Ku5htr1m One more thing, i picked up what i believe is an error on that same slide: From this: $\displaystyle \beta *l+\sigma *\left \{ \frac{\alpha -p_{x}-\sigma *l}{\beta } \right \}=\alpha -p_{l}$ To this: $\displaystyle l=\frac{\beta -\sigma }{\beta ^2-\sigma ^2}*\alpha -\frac{\beta }{\beta ^2-\sigma ^2}*p_{l}+\frac{\sigma }{\beta ^2-\sigma ^2}*p_{x}$ i believe it should be : $\displaystyle l=\frac{\beta -\sigma }{\beta -\sigma ^2}*\alpha -\frac{\beta }{\beta -\sigma ^2}*p_{l}+\frac{\sigma }{\beta -\sigma ^2}*p_{x}$
$\beta * l + \sigma * \left \{ \dfrac{\alpha - p_{x} - \sigma * l}{\beta } \right \} = \alpha - p_{l} \implies \beta^2l + \sigma ( \alpha - p_x) - \sigma^2l = \beta(\alpha - p_l) \implies$

$l( \beta^2 - \sigma^2 ) = \beta ( \alpha - p_l) - \sigma (\alpha - p_x) = \alpha \beta - \beta p_l - \alpha \sigma + \sigma p_x = ( \beta -\sigma ) \alpha - \beta p_l + \sigma p_x \implies$

$l = \dfrac{\beta - \sigma }{\beta^2 - \sigma^2} * \alpha - \dfrac{\beta}{\beta^2 - \sigma^2} * p_l + \dfrac{\sigma}{\beta^2 - \sigma^2} * p_x.$

The slide looks good to me. Are you sure you are mathematically ready for this course? September 21st, 2016, 08:01 AM   #17
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 Originally Posted by JeffM1 $\beta * l + \sigma * \left \{ \dfrac{\alpha - p_{x} - \sigma * l}{\beta } \right \} = \alpha - p_{l} \implies \beta^2l + \sigma ( \alpha - p_x) - \sigma^2l = \beta(\alpha - p_l) \implies$ $l( \beta^2 - \sigma^2 ) = \beta ( \alpha - p_l) - \sigma (\alpha - p_x) = \alpha \beta - \beta p_l - \alpha \sigma + \sigma p_x = ( \beta -\sigma ) \alpha - \beta p_l + \sigma p_x \implies$ $l = \dfrac{\beta - \sigma }{\beta^2 - \sigma^2} * \alpha - \dfrac{\beta}{\beta^2 - \sigma^2} * p_l + \dfrac{\sigma}{\beta^2 - \sigma^2} * p_x.$ The slide looks good to me. Are you sure you are mathematically ready for this course?

how did you get $\displaystyle \beta^2l$ September 21st, 2016, 08:11 AM   #18
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 Originally Posted by Ku5htr1m how did you get $\displaystyle \beta^2l$
I multiplied both sides of the equation by $\beta$ to get rid of the fraction. September 21st, 2016, 08:15 AM   #19
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 Originally Posted by JeffM1 I multiplied both sides of the equation by $\beta$ to get rid of the fraction.
ah ok i only multiplied it with the right side, i thought it was enough?

Last edited by Ku5htr1m; September 21st, 2016 at 08:22 AM. September 21st, 2016, 09:37 AM #20 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 You must not alter the value of one side of the equation without altering the other side the same way. This is basic algebra, taught in the first week or two of the subject. Seriously and intended helpfully, whatever course you are taking appears as though it demands math at a level you are very uncomfortable with. I'd drop it for now and take a college algebra course and a calculus course before trying to deal with differential equations. Tags simplification Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Chikis Elementary Math 2 September 11th, 2013 10:11 AM arron1990 Algebra 3 August 14th, 2012 04:50 AM Jakarta Algebra 3 August 4th, 2012 10:45 AM Salient Algebra 3 October 21st, 2008 01:17 PM arron1990 Calculus 1 December 31st, 1969 04:00 PM

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