The Oxford Mathematics RU-vid Channel features public lectures, student lectures, research films and more. All with the aim of explaining a subject that affects us all.
I tried that! I never did my maths homework, but I could never understand why I never passed maths! I was enthusiastically lazy! Must have been bad teaching technique!
I could ask for an explaination but I am well into life and I have survived without this knowledge. I will leave it for the next next gen who are way smarter
It’s an odd function being integrated from -pi to pi You know that for all odd functions integrated from a point -a to a point a the net signed integral will always be zero. I think that’s what he’s referring to, but I can’t be sure. It looks like a Fourier series and that’s something that does come up with them to save time so I’m making an assumption.
Yeah yeah i did exactly that i solved questions in a way that my teacher didn't like eventhough the answers were right i failed the tests and i changed the stream and it turned out great
THAT IS NOT BEING LAZY. THIS DUDE HAS A BAD HABIT OF USING THE WRONG WORDS. THE TRICK IS JUST A FACT OF THE SINE AND COSINE FUNCTIONS, A LOT OF THEM OVER π OR 2π IS ZERO UPON INTEGRATION. IT'S JUST MATHEMATICAL FACTS , NOT LAZINESS. MATHEMATICIAN HAS TO BE THE OPPOSITE OF LAZY, HE HAS TO STUDY HARD. I SAY HE BECAUSE ALL THE GREAT MATHEMATICIANS WERE MEN, NONE WERE WOMEN. NOT A SINGLE ONE.
The laziness statement isn't true. It's not working without thinking. Working without thinking even if it's more steps is less work than stepping back and thinking. It shows your stupidity because that's what stupid people think. Stupid people think that if they shit their pants they're smart because they're saving toilet paper, and having to wipe their ass. Math is expanding degrees of freedom. What you're actually doing is seeing if your steps are an actual step and expansion of a degree of freedom or an equivalent of fewer steps. You just said what stupid people like lawyers and business majors think is smart. Go back to building sand castle from where ever your from.
First fold makes a L Shape cut out of the square. Now imagine a line (let's call it c) between those two points of the L. What we get is a right triangle with a²+b² = c² (or c = √(2a²) = a*√2 ≈ 1,4a. The coaster does not fit through this diagonal of the square. Now everyone who has had Geometry in school will quickly agree that the sum of the two parts that make up this triangle (let's call them a and b, although in this spec case a = b, because its made from a square) is bigger than the line we drew. Cool, let's use this to fit the coater We fold the second time to change the angle between a and b until they become one single line, with it's length being a+b (respectively 2a, which is - surprise surprise - bigger than 1,4a).
Well cool, but in school it doesn't work like that. Without all of the proper steps you need to take to get to your solution you will get like...what... 2 out of 10 points? That's why I hated math back in my school days. It sucked like hell and I was glad when this shit was over
Is there any one who can help solve this problem, even my teacher is having problem with it , “given that h(x)=integral of ((f’(x)x-f(x))/x^2) , where 2h(2)=f(2)+4 and f(-1)=5 , what is the value of h’(-1) ? There are some steps that i have done, h(x)=f(x)/x+C and i found the value of C which is 2, thus h(x)=f(x)/x+2 and h(-1)=-3 , now how can i find the value of h’(-1) ? Appreciate your help
Is there any one who can help solve this problem, even my teacher is having problem with it , “given that h(x)=integral of ((f’(x)x-f(x))/x^2) , where 2h(2)=f(2)+4 and f(-1)=5 , what is the value of h’(-1) ? There are some steps that i have done, h(x)=f(x)/x+C and i found the value of C which is 2, thus h(x)=f(x)/x+2 and h(-1)=-3 , now how can i find the value of h’(-1) ? Appreciate your help
@@shyaamganesh9981 actually the problem isn’t about 2 expressions of h(x) coz it is okay for such things to happen since they are equall, but the problem is that there is no way finding h’(-1) given current information, it must be lacking something, the question was from a popular teacher from my country that is why I doubted if it is a mistake but i cheched with lots of people and they all said it is missing something, thanks for your help Bro, appreciate it
@@AlanNajat09 We can’t have two different expressions for h(x). What i was getting was h(x)=f(x)/x+2(which is what even you got) AND h(x)=f(x)/x. So there must be some mistake in the given conditions. Happy to help.
@@shyaamganesh9981 it is true that we can’t have 2 different expressions for h(x) but the problem is that I believe we don’t have 2 expressions, when we do the integration h(x) becomes h(x)=f(x)/x+C right? Then we plug in x=2 and plug h(2) into this equation 2h(2)=f(2)+4 in order to find the Value of C which we get C=2 , thus h(x)=f(x)/x+2 , but I don’t know how you got the second expression you mentioned where the value of C was 0 , could you please share with me how, appreciate it🤝🏻🤍
Is there any one who can help solve this problem, even my teacher is having problem with it , “given that h(x)=integral of ((f’(x)x-f(x))/x^2) , where 2h(2)=f(2)+4 and f(-1)=5 , what is the value of h’(-1) ? There are some steps that i have done, h(x)=f(x)/x+C and i found the value of C which is 2, thus h(x)=f(x)/x+2 and h(-1)=-3 , now how can i find the value of h’(-1) ? Appreciate your help
😢yeah that's why I'm unable to most mathematics questions in exams, it takes time to do them well ya know! Why do you give so less time in exams? It's not okay! 😔😭
