network profile
| bio | website | fabulastudios.co.uk |
|---|---|---|
| location | London, United Kingdom | |
| age | 19 | |
| visits | member for | 1 year, 1 month |
| seen | Dec 20 '12 at 1:42 | |
| stats | profile views | 4 |
I'm a full time student currently studying for a masters in theoretical physics at imperial college london. In my spare time I enjoy playing around with html, css, javascript, jQuery and C++. I also work and have designed the website for a small app development company fabulastudios. Check us out here
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Apr 23 |
awarded | Commentator |
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Apr 23 |
comment |
Solving a problem to do with the partial derivative chain rule. Ahhhhhhh I'm being stupid don't worry... It's neither is it? $f'(x-y) = \frac{\partial f(x-y)}{\partial s} $ ? |
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Apr 23 |
comment |
Solving a problem to do with the partial derivative chain rule. Because those are different things so when I do $ \frac{\partial^2u}{\partial x \partial y} u$ how do I notate that? |
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Apr 23 |
comment |
Solving a problem to do with the partial derivative chain rule. The only problem I have still is with your $ ' $ notation? What does that mean? is $f'(x-y) = \frac{ \partial f(x-y)}{\partial y} $ or $ \frac{ \partial f(x-y)}{\partial x} $? |
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Apr 23 |
comment |
Solving a problem to do with the partial derivative chain rule. Ahhhhh right I've got you! Didn't quite click in my mind what you meant until you showed $ \frac{\partial}{\partial y} $ Thanks, I'll have another go at it, then mark you correct :D |
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Apr 23 |
comment |
Solving a problem to do with the partial derivative chain rule. How do you find $\frac{\partial}{\partial y} u $ Using that method? or $ \frac{\partial^2}{\partial x \partial y} u $ ? |
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Apr 23 |
asked | Solving a problem to do with the partial derivative chain rule. |
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Apr 16 |
accepted | Proving formulae for Consecutive population decays. |
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Apr 13 |
answered | Proving formulae for Consecutive population decays. |
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Apr 13 |
comment |
Proving formulae for Consecutive population decays. Ok I've followed your method, however even the way your saying to do it doesn't account for the $\alpha A_0 \frac{e^{−\alpha t}−e^{−\beta t}}{\beta − \alpha}$ because you don't get the $ e^{-\beta t} $ in the numerator... :/ |
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Apr 12 |
comment |
Proving formulae for Consecutive population decays. Oooop, just realised I'm being stupid... One second! |
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Apr 12 |
comment |
Proving formulae for Consecutive population decays. This is confusing me soooo much! I've only ever done particular solutions and general solutions for second order DE's, for first order I've only ever done using integrating factors, but that doesn't work in this case because you don't know what B is, so you can't use it as an integrating factor! :/ Graghhh I'm soooo frustrated with this question :/ |
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Apr 12 |
comment |
Proving formulae for Consecutive population decays. Will I need to use an integrating factor to get the particular solution? |
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Apr 12 |
asked | Proving formulae for Consecutive population decays. |
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Apr 12 |
accepted | Evaluation of $ \lim_{ x \to \infty} x^{\frac{3}{2}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$ |
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Mar 30 |
comment |
Evaluation of $ \lim_{ x \to \infty} x^{\frac{3}{2}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$ Ahhh dammit why can't I spot these things for myself! :P |
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Mar 30 |
awarded | Supporter |
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Mar 30 |
awarded | Scholar |
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Mar 29 |
awarded | Student |
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Mar 29 |
asked | Evaluation of $ \lim_{ x \to \infty} x^{\frac{3}{2}}(\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x})$ |
